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Half-life and doubling time models

Learning goals

  • Builds exponential models from initial value and half-life or doubling time.
  • Determines half-life or doubling time from a model.
  • Solves for elapsed time using logarithms when needed.
  • Interprets parameters and predictions in context.

Click problems to add them to your worksheet.

Half-life and doubling time models 1
A sample contains \(80\,\text{mg}\) of a substance. Its half-life is \(6\) hours. How much remains after exactly one half-life?

Hints

- A half-life is the time required for an amount to become half as large. - Only one half-life has passed.

Solution

One half-life reduces the amount to one-half of its starting value. Half of \(80\,\text{mg}\) is \(40\,\text{mg}\).

Answer

\(40\,\text{mg}\)
Half-life and doubling time models 2
The graph shows the amount \(A(t)\) of a substance over time. Use the marked points and axis values to determine the half-life.
Figure for problem 551493

Hints

- Identify the initial amount from the graph. - Find a later marked amount that is exactly one half as large. - Compare the corresponding horizontal coordinates.

Solution

1. The graph shows \(A(0)=160\) and \(A(5)=80\). 2. The amount falls to one half of its initial value in \(5\) time units. 3. The next marked point, \((10, 40)\), confirms the same halving interval.

Answer

\(5\) time units
Half-life and doubling time models 3
A student-run photo page starts with \(120\) subscribers. Its simple growth model doubles the count each month. Write a formula for the number \(N(t)\) after \(t\) months.

Hints

- How can a formula show repeated doubling from the starting amount? - What does the exponent count in this model?

Solution

1. Each month multiplies the count by \(2\). 2. The exponent counts the months, so \(N(t)=120\cdot2^t\).

Answer

\(N(t)=120\cdot2^t\)
Half-life and doubling time models 4
The table follows a simple exponential growth model for an online game. Fill in the missing player count and find the doubling time. <table><tr><th>Week</th><th>Players</th></tr><tr><td>\(0\)</td><td>\(30\)</td></tr><tr><td>\(2\)</td><td>\(60\)</td></tr><tr><td>\(4\)</td><td>\(\square\)</td></tr><tr><td>\(6\)</td><td>\(240\)</td></tr></table>

Hints

- Compare the counts at week \(0\) and week \(2\). - Use the same multiplication factor in each equal time interval. - Check the final table value after filling the blank.

Solution

1. The number doubles from \(30\) to \(60\) in \(2\) weeks, so the doubling time is \(2\) weeks. 2. Double \(60\) once more for week \(4\), giving \(120\). Another doubling gives \(240\), matching week \(6\).

Answer

Missing count: \(120\) players. Doubling time: \(2\) weeks.
Half-life and doubling time models 5
A livestream's viewer count follows the graph. How long does the model take to double the viewers from \(P\) to \(Q\)?
Figure for problem 561603

Hints

- Read the two marked heights and compare them. - Find the difference between their horizontal coordinates.

Solution

1. The viewer count is \(200\) at \(P\) and \(400\) at \(Q\), so it doubles. 2. The horizontal coordinates are \(0\) and \(3\). The doubling time is \(3\) hours.

Answer

\(3\) hours.
Half-life and doubling time models 6
A tank initially contains \(48\,\text{mg}\) of dye. In a simple flushing model, half the dye remains every \(5\) minutes. Write an exponential function \(A(t)\) for the amount of dye in milligrams after \(t\) minutes.

Hints

- What amount do you have before any time has passed? - Which multiplication factor represents losing half of what remains? - How many five-minute intervals are in \(t\) minutes?

Solution

1. The starting amount is \(48\) milligrams. 2. The factor \(\frac{1}{2}\) applies once every \(5\) minutes, so the number of half-life periods is \(\frac{t}{5}\). 3. The model is \(A(t)=48\left(\frac{1}{2}\right)^{t/5}\).

Answer

\(A(t)=48\left(\frac{1}{2}\right)^{t/5}\)
Half-life and doubling time models 7
A festival's ticket waitlist is modeled by \(W(t)=500(1.08)^t\), where \(t\) is the number of days. What percent does the waitlist grow each day in this model?

Hints

- In an exponential model, what does the base tell you about one time step? - What growth factor would mean no increase at all? - Convert the part above that factor into a percentage.

Solution

1. The daily growth factor in the model is \(1.08\). 2. Subtract \(1\) to find the decimal increase: \(1.08-1=0.08\), which is \(8\%\).

Answer

\(8\%\) per day.
Half-life and doubling time models 8
A community garden uses the model \(C(t)=500-320\left(\frac{1}{2}\right)^t\) for the area covered by plants, in square feet, after \(t\) months. What area does the model approach over time?

Hints

- What happens to a number between \(0\) and \(1\) when it is raised to larger and larger powers? - Which part of the expression remains when the shrinking term gets close to zero?

Solution

1. As \(t\) increases, \(\left(\frac{1}{2}\right)^t\) approaches \(0\). 2. Therefore, \(C(t)\) approaches \(500\) square feet. The model never reaches that value at a finite time.

Answer

\(500\,\text{ft}^2\)
Half-life and doubling time models 9
A decay process is modeled by \(h(t)=250(0.75)^t\), where \(t\) is measured in seconds. 1. Rewrite the function in the form \(h(t)=h(0)e^{kt}\). Round \(k\) to four decimal places. 2. Find the half-life \(T_{1/2}\) of the process. Round to the nearest hundredth of a second. 3. Find the percent decrease per second.

Hints

- Use \(e^k=0.75\) to find \(k\). - At the half-life, the exponential factor equals \(0.5\). - Interpret the factor \(0.75\) as the percent that remains after one second.

Solution

1. The initial value is \(h(0)=250\). Since \(e^k=0.75\), \(k=\ln(0.75)\approx-0.2877\). Thus, \(h(t)=250e^{\ln(0.75)t}\approx250e^{-0.2877t}\). 2. At the half-life, \((0.75)^{T_{1/2}}=0.5\). Therefore, \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.75)}\approx2.41\,\text{s}\). 3. The factor \(0.75\) means that \(75\%\) remains each second, so the decrease is \(25\%\) per second.

Answer

1. \(h(t)=250e^{\ln(0.75)t}\approx250e^{-0.2877t}\) 2. \(T_{1/2}\approx2.41\,\text{s}\) 3. \(25\%\) per second
Half-life and doubling time models 10
The Rule of 70 estimates the doubling time \(T_d\), in years, for an investment earning \(p\%\) annually: \(pT_d\approx70\). Use the rule to estimate the doubling time for each annual rate. 1. \(1.4\%\) 2. \(2.5\%\) 3. \(3.5\%\) 4. \(5.0\%\)

Hints

- Rearrange the formula to isolate the doubling time. - Substitute each percent rate for \(p\). - The result is measured in years.

Solution

1. Solve for doubling time: \(T_d\approx\frac{70}{p}\). 2. For \(1.4\%\), \(T_d\approx\frac{70}{1.4}=50\) years. 3. For \(2.5\%\), \(T_d\approx\frac{70}{2.5}=28\) years. 4. For \(3.5\%\), \(T_d\approx\frac{70}{3.5}=20\) years. 5. For \(5.0\%\), \(T_d\approx\frac{70}{5.0}=14\) years.

Answer

1. \(50\) years 2. \(28\) years 3. \(20\) years 4. \(14\) years
Half-life and doubling time models 11
Biologist Hana Sato uses the Rule of 70, \(pT_d\approx70\), to estimate the relationship between a culture's hourly growth rate \(p\), as a percent, and its doubling time \(T_d\), in hours. Estimate the hourly growth rate for each culture. 1. Culture A: \(T_d=20\,\text{hours}\) 2. Culture B: \(T_d=35\,\text{hours}\) 3. Culture C: \(T_d=50\,\text{hours}\)

Hints

- Identify the unknown in the Rule of 70. - Consider how a longer doubling time affects the growth rate. - Substitute each doubling time into the formula.

Solution

1. Rearrange the Rule of 70: \(p\approx\frac{70}{T_d}\). 2. Culture A: \(p\approx\frac{70}{20}=3.5\), so the hourly growth rate is about \(3.5\%\). 3. Culture B: \(p\approx\frac{70}{35}=2\), so the hourly growth rate is about \(2\%\). 4. Culture C: \(p\approx\frac{70}{50}=1.4\), so the hourly growth rate is about \(1.4\%\).

Answer

1. About \(3.5\%\) per hour 2. About \(2\%\) per hour 3. About \(1.4\%\) per hour
Half-life and doubling time models 12
A forestry company estimates that the timber volume in a forest grows by \(3.5\%\) each year. The initial volume is \(2000\,\text{m}^3\). After how many years will the volume double? Round to the nearest tenth.

Hints

- Convert the annual percent increase to a growth factor. - A doubled amount is twice the initial value. - Use logarithms to solve for an unknown exponent. - Notice that the initial volume cancels.

Solution

1. The growth factor is \(1.035\). Doubling gives \(2000\cdot1.035^t=4000\). 2. Divide by \(2000\): \(1.035^t=2\). 3. Therefore, \(t=\frac{\log(2)}{\log(1.035)}\approx20.15\), which rounds to \(20.1\) years.

Answer

About \(20.1\) years
Half-life and doubling time models 13
A radioactive sample has a half-life of \(15\) minutes. At the beginning of an experiment, the sample has a mass of \(120\,\text{mg}\). a) Write a function \(f(t)=cb^t\) that gives the mass, in milligrams, after \(t\) minutes. b) Determine how many minutes it takes for the mass to decrease to \(15\,\text{mg}\).

Hints

- A half-life tells you the factor over a \(15\)-minute interval. - Express the decay using a power with base \(0.5\). - Determine how many halvings reduce \(120\,\text{mg}\) to \(15\,\text{mg}\). - Multiply the number of half-life intervals by \(15\) minutes.

Solution

1. The initial value is \(c=120\). 2. The minute-by-minute decay factor satisfies \(b^{15}=0.5\), so \(b=0.5^{1/15}\approx0.9548\). Thus \(f(t)=120\left(0.5^{1/15}\right)^t\), or equivalently \(f(t)=120(0.5)^{t/15}\). 3. Set the mass equal to \(15\): \(120(0.5)^{t/15}=15\), so \((0.5)^{t/15}=\frac{1}{8}=(0.5)^3\). 4. Therefore, \(\frac{t}{15}=3\), giving \(t=45\) minutes.

Answer

a) \(f(t)=120\left(0.5^{1/15}\right)^t\), or equivalently \(f(t)=120(0.5)^{t/15}\) b) \(45\) minutes
Half-life and doubling time models 14
A medication is eliminated from the body so that its concentration in the blood decreases by \(14\%\) each hour. Find the medication's half-life in hours. Round to the nearest hundredth.

Hints

- What fraction of the concentration remains after one hour? - Write the exponential decay factor. - At the half-life, the remaining fraction is \(0.5\). - Use logarithms to solve for time.

Solution

1. The hourly decay factor is \(0.86\). 2. At the half-life, \(0.86^t=0.5\). 3. Therefore, \(t=\frac{\ln(0.5)}{\ln(0.86)}\approx4.59577\). 4. Rounded to the nearest hundredth, the half-life is \(4.60\) hours.

Answer

About \(4.60\) hours
Half-life and doubling time models 15
Two types of algae grow in a test pond. The graphs show the area, in square meters, covered by each type after \(t\) days. Curve \(p\) represents type \(1\), and curve \(q\) represents type \(2\). a) Use the graph to determine the initial area and the doubling time for each type. b) Using the initial area and doubling time you read for type \(1\), write a model in the form \(A(t)=A_0 2^{t/T_d}\). Then find the area covered after \(10\) days. Round to the nearest hundredth of a square meter.
Figure for problem 534492

Hints

- Read the starting height of each curve at \(t=0\). - For each curve, find when the area first reaches twice its starting value. - In \(A(t)=A_0 2^{t/T_d}\), both \(A_0\) and \(T_d\) come from the graph. - After writing the type \(1\) model, substitute the requested time.

Solution

1. Both curves begin at \(5\,\text{m}^2\), so each initial area is \(5\,\text{m}^2\). 2. Twice the initial area is \(10\,\text{m}^2\). Curve \(p\) reaches \(10\,\text{m}^2\) at \(t=4\), while curve \(q\) reaches it at \(t=6\). Thus, the doubling times are \(4\) days and \(6\) days. 3. For type \(1\), \(A_0=5\) and \(T_d=4\), so \(A(t)=5\cdot2^{t/4}\). 4. \(A(10)=5\cdot2^{10/4}=5\cdot2^{2.5}\approx28.28\,\text{m}^2\).

Answer

a) Type \(1\): initial area \(5\,\text{m}^2\), doubling time \(4\) days; type \(2\): initial area \(5\,\text{m}^2\), doubling time \(6\) days b) \(A(t)=5\cdot2^{t/4}\); \(A(10)\approx28.28\,\text{m}^2\)
Half-life and doubling time models 16
A plant nursery starts with \(30\) rooted cuttings. In a simplified growth model, the number doubles every \(4\) weeks. Write a function for the predicted number \(N(t)\) after \(t\) weeks, then predict the number after \(12\) weeks.

Hints

- Use the starting number as the leading factor. - How many doubling periods have passed after \(t\) weeks? - Substitute the requested time after writing the model.

Solution

1. A doubling period is \(4\) weeks, so \(N(t)=30\cdot2^{t/4}\). 2. At \(t=12\), there are \(3\) doubling periods: \(N(12)=30\cdot2^3=240\).

Answer

\(N(t)=30\cdot2^{t/4}\); \(240\) cuttings after \(12\) weeks.
Half-life and doubling time models 17
Under an exponential decay model, a quantity is \(80\) units after \(2\) hours and \(20\) units after \(8\) hours. Find its half-life. Explain why you do not need to know its amount at time \(0\).

Hints

- What fraction of the earlier amount remains at the later time? - How many halvings produce that fraction? - Which time difference do the two observations provide?

Solution

1. Between hour \(2\) and hour \(8\), the amount falls from \(80\) to \(20\), a factor of \(\frac{1}{4}=\left(\frac{1}{2}\right)^2\). 2. Two half-life periods take \(6\) hours, so one takes \(3\) hours. 3. The decay factor between two known times depends only on the time interval, not on the starting amount.

Answer

Half-life \(3\) hours; the ratio between the two measurements does not depend on the starting amount.
Half-life and doubling time models 18
Which of the two graphed quantities has the shorter half-life? Find the half-life of each by comparing its starting height with half that height.
Figure for problem 561608

Hints

- Compare each curve with half of its own starting height, not with the other curve's height. - Locate those half-height points on the horizontal axis.

Solution

1. Curve \(f\) starts at \(8\) and reaches \(4\) after \(2\) time units, so its half-life is \(2\). 2. Curve \(g\) starts at \(4\) and reaches \(2\) after \(4\) time units, so its half-life is \(4\). 3. Curve \(f\) has the shorter half-life, even though it starts higher.

Answer

\(f\): \(2\) time units; \(g\): \(4\) time units. Curve \(f\) is shorter.
Half-life and doubling time models 19
A school store compares two predictions for daily sales in dollars. Which prediction is exponential, and what growth factor does it use each day? <table><tr><th>Day</th><th>In-store sales</th><th>Web sales</th></tr><tr><td>\(0\)</td><td>\(10\)</td><td>\(10\)</td></tr><tr><td>\(1\)</td><td>\(15\)</td><td>\(15\)</td></tr><tr><td>\(2\)</td><td>\(20\)</td><td>\(22.50\)</td></tr><tr><td>\(3\)</td><td>\(25\)</td><td>\(33.75\)</td></tr></table>

Hints

- Compare consecutive differences in each column. - Compare consecutive ratios in each column. - Which pattern is linked to exponential growth?

Solution

1. In-store sales increase by \(\$5\) each day, so their differences, not ratios, are constant. 2. Web sales are multiplied by \(\frac{15}{10}=\frac{22.5}{15}=\frac{33.75}{22.5}=1.5\) each day. 3. Therefore, the web-sales prediction is exponential, with daily factor \(1.5\).

Answer

Web sales, with growth factor \(1.5\) each day.
Half-life and doubling time models 20
A warehouse models how many packed boxes it stores. On which first whole day does the model exceed capacity? Explain why predictions beyond that point may not describe the warehouse well.
Figure for problem 561610

Hints

- How can a graph show that a predicted value has passed a fixed amount? - Which whole-day marks lie just before and after the curve first crosses the line? - Think about a real restriction the simple exponential formula ignores.

Solution

1. The capacity line is at \(1000\) boxes. The exponential curve is still below it at day \(4\) but above it at day \(5\). 2. The first whole day when the model exceeds capacity is day \(5\). 3. A warehouse cannot hold more boxes than its capacity without changing its storage or shipping process, so unrestricted exponential growth is not realistic indefinitely.

Answer

Day \(5\); the warehouse has a fixed capacity, so the model cannot describe unlimited storage growth.
Half-life and doubling time models 21
A cup of tea cools in a room kept at \(20\,{}^\circ\text{C}\). Its temperature is modeled by \(T(t)=20+36\left(\frac{1}{2}\right)^{t/4}\), with \(t\) in minutes. Find the temperature and the difference from room temperature at \(t=0\) and \(t=4\). Which quantity halves in those four minutes?

Hints

- Substitute the two times before deciding what changes by a factor of \(\frac{1}{2}\). - Compare the temperatures, then compare how far each is above room temperature.

Solution

1. \(T(0)=20+36=56\,{}^\circ\text{C}\). The difference from room temperature is \(56-20=36\,{}^\circ\text{C}\). 2. \(T(4)=20+36\cdot\frac{1}{2}=38\,{}^\circ\text{C}\). The difference is \(38-20=18\,{}^\circ\text{C}\). 3. The difference from room temperature halves, from \(36\) to \(18\). The tea's temperature does not halve.

Answer

At \(t=0\): \(56\,{}^\circ\text{C}\), difference \(36\,{}^\circ\text{C}\). At \(t=4\): \(38\,{}^\circ\text{C}\), difference \(18\,{}^\circ\text{C}\). The difference from room temperature halves.
Half-life and doubling time models 22
A greenhouse's model says the number of seedlings doubles each week. It predicts \(40\) seedlings at the end of week \(1\). How many seedlings were there at week \(0\), and what is the model \(N(t)\) for \(t\) weeks?

Hints

- If a value doubles in one week, how can you work backward one week? - Which number in an exponential model represents the value at \(t=0\)? - How should the weekly doubling factor appear in the formula?

Solution

1. Work back one doubling period: the starting number is \(40\div2=20\). 2. With a weekly factor of \(2\), the model is \(N(t)=20\cdot2^t\).

Answer

\(20\) seedlings at week \(0\); \(N(t)=20\cdot2^t\).
Half-life and doubling time models 23
A city's population follows \(P(t)=10{,}000e^{0.04t}\), where \(t\) is measured in years. Find the exact growth factor for one year and the percentage increase in one year. Round the percentage to the nearest hundredth of a percent.

Hints

- What does a growth factor tell you about the amount after a full year? - How is a growth factor turned into a percentage increase? - Is the coefficient in the exponent exactly the same as the one-year percentage increase?

Solution

1. For an increase of one year, \(\frac{P(t+1)}{P(t)}=e^{0.04}\). This is the exact growth factor. 2. The one-year percentage increase is \(100(e^{0.04}-1)\%\approx4.08\%\), not exactly \(4\%\).

Answer

Exact one-year growth factor: \(e^{0.04}\). One-year increase: about \(4.08\%\).
Half-life and doubling time models 24
A newsletter's follower count is modeled by \(F(t)=1200-800\left(\frac{1}{2}\right)^{t/4}\), with \(t\) measured in months. How many months does it take for the gap between the follower count and \(1200\) to become half as large as it was at the start?

Hints

- Is the part that is halved the total follower count or the gap below the limiting count? - Which quantity gets smaller as the follower count approaches the limiting count? - What exponent makes the factor \(\frac{1}{2}\) appear once?

Solution

1. The gap is \(1200-F(t)=800\left(\frac{1}{2}\right)^{t/4}\). 2. The gap is halved when the exponent is \(1\), so \(t/4=1\) and \(t=4\) months.

Answer

\(4\) months.
Half-life and doubling time models 25
A park estimates that a bird population follows \(N(t)=80(1.5)^t\), where \(t\) is the number of years since the study began. At which first whole-number year does the model predict more than \(250\) birds?

Hints

- Check the model at neighboring whole-number years rather than rounding an earlier count upward. - What does “more than” mean at the threshold? - Does this model grow or shrink as time increases?

Solution

1. At year \(2\), the model gives \(N(2)=80(1.5)^2=180\), which is below \(250\). 2. At year \(3\), it gives \(N(3)=80(1.5)^3=270\), which exceeds \(250\). The count rises every year, so year \(3\) is the first whole-number year above the threshold.

Answer

Year \(3\).
Half-life and doubling time models 26
A savings account follows \(B(t)=300(1.10)^{2t}\), where \(t\) is measured in years. How often is the \(10\%\) increase applied, and what is the effective percentage increase over one year?

Hints

- How many times does the stated factor occur when \(t=1\)? - Which time interval makes the exponent increase by one? - How do you convert a one-year growth factor into a percent increase?

Solution

1. The exponent is \(2t\), so the factor \(1.10\) is applied twice per year: once every six months. 2. The one-year factor is \((1.10)^2=1.21\), so the effective yearly increase is \(21\%\).

Answer

Every six months; an effective yearly increase of \(21\%\).
Half-life and doubling time models 27
The Cedar robotics newsletter has \(A(t)=64\cdot2^{t/2}\) subscribers in its model, and the Maple chess newsletter has \(B(t)=16\cdot2^t\) subscribers, \(t\) months after they launch. When do the models predict the same number of subscribers? Why does the chess newsletter catch up although it starts smaller?

Hints

- What equation expresses equal subscriber counts at the same time? - Can both exponential parts be written with base \(2\)? - Compare the doubling intervals, not just the starting numbers.

Solution

1. Set the models equal: \(64\cdot2^{t/2}=16\cdot2^t\). 2. Divide by \(16\cdot2^{t/2}\) to get \(4=2^{t/2}\). Since \(4=2^2\), we have \(t/2=2\), so \(t=4\) months. 3. The chess newsletter starts with fewer subscribers, but its model doubles every month. The robotics newsletter doubles every two months, so the chess newsletter catches up.

Answer

After \(4\) months. The chess newsletter doubles every month, while the robotics newsletter doubles every two months.
Half-life and doubling time models 28
Hydrogen peroxide (\(\text{H}_2\text{O}_2\)) decays exponentially in the presence of a catalyst. During an experiment, half of the hydrogen peroxide decays in the first \(10\) minutes. How many minutes after the experiment begins will \(90\%\) of the hydrogen peroxide have decayed? a) \(18\,\text{min}\) b) \(30\,\text{min}\) c) \(32.1\,\text{min}\) d) \(33.2\,\text{min}\)

Hints

- Use the half-life to write an exponential decay model. - If \(90\%\) has decayed, what percent remains? - Set the remaining fraction equal to the decay model, then use logarithms to solve for time.

Solution

1. With a half-life of \(10\) minutes, the amount remaining is \(N(t)=N_0\cdot0.5^{\frac{t}{10}}\). 2. If \(90\%\) has decayed, then \(10\%\) remains, so \(0.1N_0=N_0\cdot0.5^{\frac{t}{10}}\). 3. Divide by \(N_0\): \(0.1=0.5^{\frac{t}{10}}\). 4. Take natural logarithms: \(\ln(0.1)=\frac{t}{10}\ln(0.5)\). 5. Therefore, \(t=10\cdot\frac{\ln(0.1)}{\ln(0.5)}\approx33.2\,\text{min}\).

Answer

d) \(33.2\,\text{min}\)
Half-life and doubling time models 29
A medication is eliminated from the body over time. Immediately after the dose is taken (\(t=0\)), its concentration in the blood is \(80\,\text{mg/L}\). After \(4\) hours, the concentration has decreased to \(50\,\text{mg/L}\). 1. Find the hourly decay factor \(a\) for a model of the form \(c(t)=c_0a^t\). Round \(a\) to four decimal places. 2. Rewrite the model in the form \(c(t)=c_0e^{kt}\). Round \(k\) to four decimal places. 3. Find the half-life \(T_{1/2}\) of the medication. Round to the nearest tenth of an hour.

Hints

- Substitute the two given concentrations into the exponential model. - In a decay model, should \(k\) be positive or negative? - Half-life is the time required for the concentration to reach half its initial value. - Set the model equal to one-half of the initial concentration and solve for time.

Solution

1. Since \(c(0)=80\), the initial concentration is \(c_0=80\). Using \(c(4)=50\) gives \(80a^4=50\), so \(a^4=\frac{5}{8}\). Therefore, \(a=\left(\frac{5}{8}\right)^{1/4}\approx0.8891\). 2. Because \(a=e^k\), \(k=\ln(a)=\frac{1}{4}\ln\left(\frac{5}{8}\right)\approx-0.1175\). Thus, \(c(t)=80e^{\frac{1}{4}\ln\left(\frac{5}{8}\right)t}\approx80e^{-0.1175t}\). 3. At the half-life, \(e^{kT_{1/2}}=0.5\). Therefore, \(T_{1/2}=\frac{\ln(0.5)}{k}\approx5.90\) hours, which rounds to \(5.9\) hours.

Answer

1. \(a\approx0.8891\) 2. \(c(t)\approx80e^{-0.1175t}\) 3. \(T_{1/2}\approx5.9\,\text{hours}\)
Half-life and doubling time models 30
A growth process is modeled by \(p(t)=15(1.04)^{0.5t}\), where \(t\) is measured in seconds. 1. Rewrite the function in the form \(p(t)=p(0)e^{kt}\). Round \(k\) to four decimal places. 2. Find the doubling time \(T_d\). Round to the nearest hundredth of a second. 3. Find the percent increase per second. Round to the nearest hundredth of a percent.

Hints

- Use exponent rules to rewrite \((1.04)^{0.5t}\) with \(t\) as the entire variable exponent. - Use \(a^t=e^{\ln(a)t}\). - At the doubling time, the exponential factor equals \(2\). - Find the growth factor for exactly one second before converting it to a percent.

Solution

1. The initial value is \(p(0)=15\). Since \((1.04)^{0.5t}=e^{0.5\ln(1.04)t}\), \(k=0.5\ln(1.04)\approx0.0196\). Therefore, \(p(t)=15e^{0.5\ln(1.04)t}\approx15e^{0.0196t}\). 2. At the doubling time, \((1.04)^{0.5T_d}=2\). Thus, \(T_d=\frac{2\ln2}{\ln(1.04)}\approx35.35\,\text{s}\). 3. The one-second growth factor is \((1.04)^{0.5}\approx1.0198\), so the increase is about \(1.98\%\) per second.

Answer

1. \(p(t)=15e^{0.5\ln(1.04)t}\approx15e^{0.0196t}\) 2. \(T_d\approx35.35\,\text{s}\) 3. About \(1.98\%\) per second
Half-life and doubling time models 31
Find the doubling time \(T_d\) or half-life \(T_{1/2}\) for each function. Round each answer to two decimal places. a) \(f(t)=40(1.15)^t\) b) \(g(t)=100e^{-0.5t}\) c) \(h(t)=15(0.98)^t\)

Hints

- First decide whether each function represents growth or decay. - At a doubling time, the exponential factor equals \(2\); at a half-life, it equals \(0.5\). - The initial coefficient does not affect the doubling time or half-life.

Solution

1. For \(f\), solve \((1.15)^{T_d}=2\). Thus, \(T_d=\frac{\ln2}{\ln(1.15)}\approx4.96\). 2. For \(g\), solve \(e^{-0.5T_{1/2}}=0.5\). Thus, \(T_{1/2}=\frac{\ln(0.5)}{-0.5}\approx1.39\). 3. For \(h\), solve \((0.98)^{T_{1/2}}=0.5\). Thus, \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.98)}\approx34.31\).

Answer

a) \(T_d\approx4.96\) b) \(T_{1/2}\approx1.39\) c) \(T_{1/2}\approx34.31\)
Half-life and doubling time models 32
A quantity is modeled by \(f(t)=f(0)e^{kt}\), where \(t\) is measured in hours. a) One quantity doubles every \(12\) hours. Find its growth coefficient \(k\). Round to four decimal places. b) Another quantity has a half-life of \(25\) hours. Find the time when only \(10\%\) of its initial amount remains. Round to the nearest hundredth of an hour.

Hints

- Relate \(k\) to the given doubling time. - Use the half-life to determine the decay coefficient for part b). - When \(10\%\) remains, the ratio \(f(t)/f(0)\) equals \(0.10\).

Solution

1. For a doubling time \(T_d=12\) hours, \(e^{12k}=2\). Therefore, \(k=\frac{\ln2}{12}\approx0.0578\). 2. A half-life \(T_{1/2}=25\) hours gives \(e^{25k}=0.5\), so \(k=\frac{\ln(0.5)}{25}\). 3. To find when \(10\%\) remains, solve \(e^{kt}=0.10\). Thus, \(t=\frac{\ln(0.10)}{k}=\frac{25\ln(0.10)}{\ln(0.5)}\approx83.05\) hours.

Answer

a) \(k\approx0.0578\) b) \(t\approx83.05\,\text{hours}\)
Half-life and doubling time models 33
A new app has \(5000\) users at \(t=0\), where \(t\) is measured in months. After \(3\) months, the app has \(8000\) users. Valeria claims that the number of users grows linearly. Diego claims that it grows exponentially and doubles every \(5\) months. After \(6\) months, the app has \(12{,}800\) users. a) Determine which model type fits the data. b) Evaluate Diego's claim about the doubling time. Round the computed doubling time to the nearest hundredth of a month.

Hints

- Compare differences and ratios over the two equal \(3\)-month intervals. - A doubling time satisfies an equation in which the growth factor equals \(2\). - Use logarithms to solve for the unknown time.

Solution

1. In the first \(3\)-month interval, the increase is \(8000-5000=3000\). In the second interval, the increase is \(12{,}800-8000=4800\). The differences are not equal, so a linear model does not fit. 2. The growth factors are \(\frac{8000}{5000}=1.6\) and \(\frac{12{,}800}{8000}=1.6\). Since the factors are equal over equal time intervals, an exponential model fits. 3. If \(T_d\) is the doubling time, then \((1.6)^{T_d/3}=2\). Therefore, \(T_d=\frac{3\ln2}{\ln(1.6)}\approx4.42\) months. 4. Diego is correct about the exponential model type but incorrect about the claimed \(5\)-month doubling time.

Answer

a) Exponential growth fits the data. b) \(T_d\approx4.42\) months, not \(5\) months.
Half-life and doubling time models 34
A medication is eliminated from the body. At \(t=0\), the body contains \(250\,\text{mg}\) of the medication. The amount decreases by \(15\%\) each hour, and \(t\) is measured in hours. a) Write a model of the form \(m(t)=ab^t\). b) Rewrite the model in the form \(m(t)=ae^{kt}\). Round \(k\) to four decimal places. c) Find the half-life \(T_{1/2}\) of the medication. Round to the nearest hundredth of an hour. d) Find when only \(10\,\text{mg}\) remains. Round to the nearest hundredth of an hour.

Hints

- Convert the percent decrease to a multiplication factor. - Use \(e^k=b\) to rewrite the model with base \(e\). - At the half-life, the ratio of the remaining amount to the initial amount is \(0.5\). - Use logarithms when the unknown is in the exponent.

Solution

1. A \(15\%\) hourly decrease gives the hourly factor \(b=1-0.15=0.85\). Thus, \(m(t)=250(0.85)^t\). 2. Since \(e^k=0.85\), \(k=\ln(0.85)\approx-0.1625\). Therefore, \(m(t)=250e^{\ln(0.85)t}\approx250e^{-0.1625t}\). 3. For the half-life \(T_{1/2}\), solve \((0.85)^{T_{1/2}}=0.5\). Thus, \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.85)}\approx4.27\) hours. 4. To find when \(10\,\text{mg}\) remains, solve \(250(0.85)^t=10\). Then \((0.85)^t=0.04\), so \(t=\frac{\ln(0.04)}{\ln(0.85)}\approx19.81\) hours.

Answer

a) \(m(t)=250(0.85)^t\) b) \(m(t)=250e^{\ln(0.85)t}\approx250e^{-0.1625t}\) c) \(T_{1/2}\approx4.27\,\text{hours}\) d) About \(19.81\,\text{hours}\)
Half-life and doubling time models 35
The population of a city is modeled by \(P(t)=45{,}000e^{0.028t}\), where \(t\) is the number of years since 2020. a) Give the population at the beginning of 2020. b) Find the annual growth factor to four decimal places and the annual percent growth rate to the nearest hundredth of a percent. c) Determine during which year the city's population first exceeds \(75{,}000\). d) Find the amount of time required for the population to triple. Round to the nearest hundredth of a year.

Hints

- Interpret the coefficient in front of the exponential expression. - The one-year factor is \(e^{0.028}\). - Use logarithms to solve for a time in an exponential equation. - The initial value cancels when finding a tripling time.

Solution

1. At \(t=0\), \(P(0)=45{,}000\). 2. The annual growth factor is \(b=e^{0.028}\approx1.0284\). Therefore, the annual growth rate is about \(2.84\%\). 3. Solve \(45{,}000e^{0.028t}=75{,}000\). This gives \(e^{0.028t}=\frac{5}{3}\), so \(t=\frac{\ln(5/3)}{0.028}\approx18.24\). The population exceeds \(75{,}000\) during 2038. 4. For the tripling time, solve \(e^{0.028t}=3\). Thus, \(t=\frac{\ln3}{0.028}\approx39.24\) years.

Answer

a) \(45{,}000\) people b) Growth factor: about \(1.0284\); growth rate: about \(2.84\%\) c) During 2038 d) About \(39.24\,\text{years}\)
Half-life and doubling time models 36
A forestry district tracks two types of timber. The current spruce volume is \(8000\,\text{m}^3\) and decreases by \(2.8\%\) per year. The current beech volume is \(2500\,\text{m}^3\) and increases by \(1.5\%\) per year because of reforestation. a) Write an exponential model of the form \(V(t)=V_0e^{kt}\) for each type of timber, where \(t\) is measured in years. b) Find the half-life \(T_{1/2}\) of the spruce volume. Round to the nearest hundredth of a year. c) Find when the beech volume reaches \(4000\,\text{m}^3\). Round to the nearest hundredth of a year. d) Find when the two timber volumes are equal. Round to the nearest hundredth of a year.

Hints

- Convert each annual percent change to a yearly factor and use \(k=\ln(q)\). - The initial value cancels when finding a half-life. - Use logarithms to solve equations with time in the exponent. - Set the two models equal to find when the timber volumes match.

Solution

1. For spruce, the yearly factor is \(0.972\), so \(k_S=\ln(0.972)\approx-0.0284\). Thus, \(V_S(t)=8000e^{\ln(0.972)t}\). 2. For beech, the yearly factor is \(1.015\), so \(k_B=\ln(1.015)\approx0.0149\). Thus, \(V_B(t)=2500e^{\ln(1.015)t}\). 3. For the spruce half-life, solve \(e^{k_ST_{1/2}}=0.5\). Then \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.972)}\approx24.41\) years. 4. For beech to reach \(4000\,\text{m}^3\), solve \(2500e^{k_Bt}=4000\). Then \(t=\frac{\ln(1.6)}{\ln(1.015)}\approx31.57\) years. 5. Set the volumes equal: \(8000e^{k_St}=2500e^{k_Bt}\). Thus, \(t=\frac{\ln(8000/2500)}{k_B-k_S}\approx26.87\) years.

Answer

a) \(V_S(t)=8000e^{\ln(0.972)t}\); \(V_B(t)=2500e^{\ln(1.015)t}\) b) \(T_{1/2}\approx24.41\,\text{years}\) c) About \(31.57\,\text{years}\) d) About \(26.87\,\text{years}\)
Half-life and doubling time models 37
Registrations for a workshop are modeled by \(g(t)=120-90e^{-0.2t}\), where \(t\ge0\) is measured in days. a) Find all starting times \(t\) for which the number of registrations can double in a finite amount of time. b) Find the doubling time when the starting time is \(t=0\). c) Explain how the required doubling time changes as the starting time \(t\) increases.

Hints

- Identify the limiting value of the function. - Which starting values could still double without reaching the model's upper limit? - How can you use the model to describe a future registration count twice as large as today's? - Consider how the growth rate changes as the function approaches its limit.

Solution

1. Registrations approach but never reach the limiting value \(120\). A future doubling is possible only when \(2g(t)<120\), or \(g(t)<60\). 2. Solve \(120-90e^{-0.2t}<60\). This gives \(e^{-0.2t}>\frac{2}{3}\), so \(t<5\ln(1.5)\approx2.03\). Thus, doubling is possible for \(0\le t<5\ln(1.5)\). 3. At \(t=0\), \(g(0)=30\). The doubling time satisfies \(g(T_d)=60\), so \(e^{-0.2T_d}=\frac{2}{3}\). Hence, \(T_d=5\ln(1.5)\approx2.03\) days. 4. As the starting time increases, registrations are closer to their limiting number and grow more slowly. The time needed to double increases without bound as \(t\) approaches \(5\ln(1.5)\).

Answer

a) \(0\le t<5\ln(1.5)\approx2.03\) b) \(T_d=5\ln(1.5)\approx2.03\,\text{days}\) c) For starting times \(0\le t<5\ln(1.5)\), the doubling time increases as \(t\) increases. At later starting times, doubling is not possible in a finite time.
Half-life and doubling time models 38
A microorganism population grows continuously according to \(N(t)=N_0e^{kt}\), where \(t\) is measured in hours. At \(t=0\), the population is \(1200\). After \(3\) hours, it is \(1850\). a) Find \(k\), rounded to four decimal places, and write the population model. b) Find the percent increase during the first hour. Round to the nearest hundredth of a percent. c) Find the doubling time \(T_d\) of the initial population. Round to the nearest hundredth of an hour.

Hints

- Substitute the value at \(t=3\) into the model and use a logarithm. - The one-hour growth factor is \(e^k\). - At the doubling time, the exponential factor equals \(2\).

Solution

1. Using \(N(3)=1850\), \(1850=1200e^{3k}\). Therefore, \(k=\frac{1}{3}\ln\left(\frac{1850}{1200}\right)\approx0.1443\), so \(N(t)=1200e^{\frac{1}{3}\ln\left(\frac{1850}{1200}\right)t}\approx1200e^{0.1443t}\). 2. The one-hour growth factor is \(e^k\approx1.1552\), so the population increases by about \(15.52\%\) during the first hour. 3. For the doubling time \(T_d\), solve \(e^{kT_d}=2\). Thus, \(T_d=\frac{\ln2}{k}\approx4.80\) hours.

Answer

a) \(k\approx0.1443\); \(N(t)=1200e^{\frac{1}{3}\ln\left(\frac{1850}{1200}\right)t}\approx1200e^{0.1443t}\) b) About \(15.52\%\) c) \(T_d\approx4.80\,\text{hours}\)
Half-life and doubling time models 39
An investment grows by a fixed percentage each year and is modeled by \(K(t)=K_0q^t\), where \(t\) is time in years and \(q\) is the growth factor. The time needed for the initial investment \(K_0\) to double is the doubling time \(T_d\). a) Show that \(T_d=\frac{\log(2)}{\log(q)}\). b) Find the doubling time for an investment earning \(4.5\%\) interest per year. Round to the nearest hundredth of a year.

Hints

- What value must the investment have at the doubling time? - How can you solve an equation when the unknown appears in an exponent? - Convert the annual percent increase to a growth factor. - First divide both sides by the common factor \(K_0\).

Solution

1. At the doubling time, \(K(T_d)=2K_0\), so \(2K_0=K_0q^{T_d}\). 2. Divide by \(K_0\): \(2=q^{T_d}\). 3. Take logarithms: \(\log(2)=T_d\log(q)\). Therefore, \(T_d=\frac{\log(2)}{\log(q)}\). 4. A yearly interest rate of \(4.5\%\) gives \(q=1+0.045=1.045\). 5. Thus, \(T_d=\frac{\log(2)}{\log(1.045)}\approx15.75\) years.

Answer

a) \(T_d=\frac{\log(2)}{\log(q)}\) b) About \(15.75\) years
Half-life and doubling time models 40
Two radioactive isotopes decay at different rates. Isotope A is modeled by \(N_A(t)=N_0\cdot0.92^t\), where \(t\) is measured in days. Isotope B has a half-life of exactly \(5\) days. a) Find the half-life \(T_{1/2}\) of Isotope A to the nearest hundredth of a day. b) Find the daily decay factor \(b\) for Isotope B if its decay is modeled by \(N_B(t)=N_0b^t\). Round \(b\) to three decimal places.

Hints

- At the half-life, what fraction of the initial amount remains? - How can you solve when the unknown appears in an exponent? - How can you solve \(b^5=0.5\) when the unknown is the base? - Keep track of which information belongs to each isotope.

Solution

1. At the half-life, \(0.5N_0=N_0\cdot0.92^{T_{1/2}}\), so \(0.5=0.92^{T_{1/2}}\). 2. Take logarithms: \(\log(0.5)=T_{1/2}\log(0.92)\). Therefore, \(T_{1/2}=\frac{\log(0.5)}{\log(0.92)}\approx8.31\) days. 3. For Isotope B, \(0.5=b^5\). 4. Taking the fifth root gives \(b=0.5^{1/5}\approx0.871\).

Answer

a) \(T_{1/2}\approx8.31\,\text{days}\) b) \(b\approx0.871\)
Half-life and doubling time models 41
The Rule of 70 estimates the doubling time \(T_d\), in years, for an investment growing at a constant annual rate of \(p\%\): \(T_d\approx\frac{70}{p}\). a) Use the Rule of 70 to estimate the doubling time for annual rates of \(2\%\), \(5\%\), and \(10\%\). b) Find the exact doubling time for each rate using logarithms. Round to the nearest hundredth of a year. c) For which rate does the Rule of 70 have the smallest error?

Hints

- A doubling means the growth factor raised to the unknown time equals \(2\). - Use logarithms to solve for an exponent. - Convert each percent rate to a growth factor. - Compare the absolute differences between each estimate and exact value.

Solution

1. Using \(T_d\approx\frac{70}{p}\): for \(2\%\), \(T_d\approx35\) years; for \(5\%\), \(T_d\approx14\) years; for \(10\%\), \(T_d\approx7\) years. 2. The exact doubling time satisfies \(\left(1+\frac{p}{100}\right)^{T_d}=2\), so \(T_d=\frac{\ln2}{\ln\left(1+\frac{p}{100}\right)}\). 3. For \(2\%\), \(T_d\approx\frac{\ln2}{\ln(1.02)}\approx35.00\) years. 4. For \(5\%\), \(T_d\approx\frac{\ln2}{\ln(1.05)}\approx14.21\) years. 5. For \(10\%\), \(T_d\approx\frac{\ln2}{\ln(1.10)}\approx7.27\) years. 6. The absolute errors are about \(0.00\), \(0.21\), and \(0.27\) years, respectively. The smallest error occurs at \(2\%\).

Answer

a) \(35\) years, \(14\) years, and \(7\) years b) \(35.00\) years, \(14.21\) years, and \(7.27\) years c) \(2\%\)
Half-life and doubling time models 42
Under ideal conditions, a bacterial culture doubles every \(14\) hours. a) Use the Rule of 70, \(pT_d\approx70\), to estimate the hourly growth rate \(p\) as a percent. b) Find the exact hourly growth factor \(b\) and the corresponding growth rate. Round the factor to four decimal places and the rate to two decimal places. c) The culture begins with \(500\) bacteria. Use the exact factor to predict the population after exactly \(3\) days. Round to the nearest whole bacterium.

Hints

- How can you rearrange the Rule of 70 to estimate an hourly rate? - What equation relates the one-hour factor to doubling in fourteen hours? - How many hours are there in three days? - Use the exact base for the prediction and round only at the end.

Solution

1. Using \(14p\approx70\), \(p\approx\frac{70}{14}=5\), so the estimated hourly rate is \(5\%\). 2. The exact factor satisfies \(b^{14}=2\), so \(b=2^{1/14}\approx1.0507566\). Thus, the hourly rate is \((b-1)\cdot100\%\approx5.08\%\). 3. Three days is \(72\) hours. The population is \(500b^{72}=500\cdot2^{72/14}\approx17{,}665.43\), or about \(17{,}665\) bacteria.

Answer

a) About \(5\%\) b) \(b\approx1.0508\); growth rate about \(5.08\%\) c) About \(17{,}665\) bacteria
Half-life and doubling time models 43
An amount of \(\$4000.00\) is invested at a fixed annual interest rate of \(1.5\%\), compounded annually. a) Find the balance after \(8\) years. Round to the nearest cent. b) After how many complete years will the balance first reach or exceed \(\$5000.00\)? c) Find the doubling time in two ways: using logarithms and using the Rule of 70, \(T_d\approx\frac{70}{p}\). Round both time estimates to the nearest hundredth of a year.

Hints

- Convert the percent increase to a growth factor. - Use logarithms to locate the threshold, then respect the annual compounding checkpoints. - Why does the starting balance cancel when finding doubling time? - In the Rule of 70, use the numerical percent rate without the percent sign.

Solution

1. The growth factor is \(1.015\), so \(K(n)=4000\cdot1.015^n\). 2. \(K(8)=4000\cdot1.015^8\approx4505.97\). The balance is about \(\$4505.97\). 3. Solve \(4000\cdot1.015^n=5000\). Then \(n=\frac{\ln(1.25)}{\ln(1.015)}\approx14.99\). Because interest is compounded annually, the first complete-year checkpoint at or above \(\$5000.00\) is \(n=15\). 4. For the logarithmic doubling-time calculation, solve \(1.015^{T_d}=2\): \(T_d=\frac{\ln2}{\ln(1.015)}\approx46.56\) years. 5. The Rule of 70 gives \(T_d\approx\frac{70}{1.5}\approx46.67\) years.

Answer

a) About \(\$4505.97\) b) \(15\) complete years c) Logarithms: about \(46.56\) years; Rule of 70: about \(46.67\) years
Half-life and doubling time models 44
A bacterial culture begins with \(250\) bacteria and grows by \(8\%\) each hour. a) State the hourly growth factor \(q\) and write a function \(B(t)\) for the population after \(t\) hours. b) After how many hours will the population have quadrupled? Round to the nearest hundredth of an hour. c) Find the doubling time \(T_d\) and compare it with the Rule of 70 estimate, \(T_d\approx\frac{70}{p}\). Round the exact doubling time to the nearest hundredth of an hour.

Hints

- Convert the hourly percent increase to a growth factor. - A quadrupled population is four times the initial population. - The starting value cancels when finding doubling time. - Compare the exact result and the Rule of 70 estimate by subtracting.

Solution

1. The growth factor is \(q=1+0.08=1.08\), so \(B(t)=250\cdot1.08^t\). 2. For a quadrupled population, solve \(250\cdot1.08^t=1000\), so \(1.08^t=4\). 3. Therefore, \(t=\frac{\ln4}{\ln(1.08)}\approx18.01\) hours. 4. The exact doubling time satisfies \(1.08^{T_d}=2\), so \(T_d=\frac{\ln2}{\ln(1.08)}\approx9.01\) hours. 5. The Rule of 70 gives \(T_d\approx\frac{70}{8}=8.75\) hours. The estimate is about \(0.26\) hour, or about \(15\) minutes, shorter than the exact value.

Answer

a) \(q=1.08\); \(B(t)=250\cdot1.08^t\) b) About \(18.01\) hours c) Exact: \(T_d\approx9.01\) hours; Rule of 70: \(8.75\) hours
Half-life and doubling time models 45
A rare wild-bee population in a protected area grows by \(6.5\%\) each year. a) After how many years will the population double? Round to the nearest tenth of a year. b) The habitat can support at most four times the current population. After how many years will the population reach that capacity if the growth rate remains constant? Round to the nearest tenth of a year.

Hints

- Convert the percent increase to a yearly growth factor. - Write an equation in which the growth factor raised to time equals \(2\). - Use logarithms to solve for the exponent. - How is quadrupling related to two consecutive doublings?

Solution

1. The yearly growth factor is \(1.065\). 2. For doubling, solve \(1.065^t=2\). Thus, \(t=\frac{\log2}{\log(1.065)}\approx11.01\) years, which rounds to \(11.0\) years. 3. For quadrupling, solve \(1.065^t=4\). Thus, \(t=\frac{\log4}{\log(1.065)}\approx22.01\) years, which rounds to \(22.0\) years. 4. Since \(4=2^2\), the quadrupling time is twice the doubling time.

Answer

a) About \(11.0\) years b) About \(22.0\) years
Half-life and doubling time models 46
A manufacturing company increases its annual production by \(5.2\%\) each year. At the same time, its material cost per unit increases by \(2.8\%\) each year. Find the doubling time \(T_d\), in years, for each quantity, assuming the rates remain constant. Round each to the nearest tenth of a year.

Hints

- Convert each percent increase to a growth factor. - Set the growth factor raised to time equal to \(2\). - The initial values cancel when finding doubling time.

Solution

1. The growth factors are \(1.052\) for production and \(1.028\) for material cost. 2. For production, solve \(1.052^{T_d}=2\): \(T_d=\frac{\log2}{\log(1.052)}\approx13.67\) years, or \(13.7\) years to the nearest tenth. 3. For material cost, solve \(1.028^{T_d}=2\): \(T_d=\frac{\log2}{\log(1.028)}\approx25.10\) years, or \(25.1\) years to the nearest tenth.

Answer

Production: \(T_d\approx13.7\,\text{years}\) Material cost per unit: \(T_d\approx25.1\,\text{years}\)
Half-life and doubling time models 47
The rule of thumb \(p\cdot T_{1/2}\approx70\) can estimate the half-life \(T_{1/2}\) of an exponential decay process with a decrease rate of \(p\%\) per time unit. a) A radioactive sample loses \(10\%\) of its activity each hour. Estimate its half-life using the rule of thumb. b) Find the exact half-life by solving \(\left(1-\frac{p}{100}\right)^{T_{1/2}}=0.5\). Round to the nearest hundredth of an hour. c) Find the percent error of the estimate relative to the exact value. Round to the nearest tenth of a percent.

Hints

- A percent decrease produces a decay factor less than \(1\). - At the half-life, one half of the initial amount remains. - Use the exact value in the denominator when calculating percent error.

Solution

1. Using \(p=10\), \(T_{1/2}\approx\frac{70}{10}=7\) hours. 2. Solve \(0.9^{T_{1/2}}=0.5\): \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.9)}\approx6.58\) hours. 3. The percent error is \(\frac{|7-6.5788|}{6.5788}\cdot100\%\approx6.40\%\), which rounds to \(6.4\%\).

Answer

a) \(7\) hours b) \(T_{1/2}\approx6.58\,\text{hours}\) c) About \(6.4\%\)
Half-life and doubling time models 48
An investment earns a fixed annual interest rate and doubles in exactly \(14\) years. a) Find the annual interest rate \(p\). Round to the nearest hundredth of a percent. b) Find how long it takes for the original investment to grow to ten times its initial value. Round to the nearest tenth of a year.

Hints

- Write an equation relating the growth factor to the \(14\)-year doubling time. - Convert the growth factor to a percent rate. - The initial investment cancels when working with a growth multiple. - For tenfold growth, set the growth factor raised to time equal to \(10\). - Use logarithm properties to simplify the expression for time.

Solution

1. If \(q\) is the annual growth factor, then \(q^{14}=2\), so \(q=2^{1/14}\approx1.05076\). 2. Therefore, \(p=(q-1)\cdot100\%\approx5.08\%\). 3. For tenfold growth, \(q^t=10\). 4. Thus, \(t=\frac{\ln10}{\ln(q)}=14\cdot\frac{\ln10}{\ln2}\approx46.51\) years, which rounds to \(46.5\) years.

Answer

a) About \(5.08\%\) b) About \(46.5\) years
Half-life and doubling time models 49
Two radioactive isotopes decay exponentially. Isotope A has a half-life of \(8\) hours, and isotope B has a half-life of \(14\) hours. Find the hourly percent decrease for each isotope. Round to the nearest hundredth of a percent.

Hints

- After one half-life, one-half of the original amount remains. - Write an equation in which the hourly decay factor is unknown. - Relate the decay factor to the percent decrease. - Use a root or a fractional exponent to solve for the factor.

Solution

1. If \(q\) is the hourly decay factor and \(T_{1/2}\) is the half-life, then \(q^{T_{1/2}}=0.5\), so \(q=0.5^{1/T_{1/2}}\). 2. For isotope A, \(q=0.5^{1/8}\approx0.9170\). The hourly percent decrease is \((1-q)\cdot100\%\approx8.30\%\). 3. For isotope B, \(q=0.5^{1/14}\approx0.9517\). The hourly percent decrease is \((1-q)\cdot100\%\approx4.83\%\).

Answer

Isotope A: about \(8.30\%\) per hour Isotope B: about \(4.83\%\) per hour
Half-life and doubling time models 50
The area of a lake covered by algae grows exponentially. During a cool summer, the covered area doubles every \(10\) days. During a very warm summer, it doubles every \(6\) days. Find the daily percent growth rate for each summer. Round to the nearest hundredth of a percent.

Hints

- A doubling means the total growth factor over the given time is \(2\). - Write an equation of the form \(q^{T_d}=2\). - Use a root or a fractional exponent to solve for \(q\). - Convert the growth factor to a percent growth rate.

Solution

1. If \(q\) is the daily growth factor and \(T_d\) is the doubling time, then \(q^{T_d}=2\), so \(q=2^{1/T_d}\). 2. For the cool summer, \(q=2^{1/10}\approx1.0718\). The daily growth rate is \((q-1)\cdot100\%\approx7.18\%\). 3. For the very warm summer, \(q=2^{1/6}\approx1.1225\). The daily growth rate is \((q-1)\cdot100\%\approx12.25\%\).

Answer

Cool summer: about \(7.18\%\) per day Very warm summer: about \(12.25\%\) per day
Half-life and doubling time models 51
For exponential growth at a rate of \(p\%\) per time unit, the time \(n\) needed for a quantity to triple can be estimated by \(n\approx\frac{110}{p}\). 1. For annual rates of \(4\%\) and \(8\%\), find the exact tripling time to the nearest hundredth of a year. 2. Use the rule of thumb for each rate and find the error compared with the exact value.

Hints

- Write an exponential growth equation for a tripled quantity. - Convert each percent increase to a growth factor. - Use logarithms to solve an equation of the form \(q^n=3\). - Compare each estimate with its corresponding exact value.

Solution

1. The exact tripling time satisfies \(\left(1+\frac{p}{100}\right)^n=3\), so \(n=\frac{\ln(3)}{\ln\left(1+\frac{p}{100}\right)}\). 2. For \(p=4\), \(n\approx\frac{\ln(3)}{\ln(1.04)}\approx28.01\) years. 3. For \(p=8\), \(n\approx\frac{\ln(3)}{\ln(1.08)}\approx14.27\) years. 4. The estimates are \(\frac{110}{4}=27.50\) years and \(\frac{110}{8}=13.75\) years. 5. The errors are about \(28.01-27.50=0.51\) year and \(14.27-13.75=0.52\) year. The rule slightly underestimates both times.

Answer

1. At \(4\%\): \(28.01\) years; at \(8\%\): \(14.27\) years 2. Estimates: \(27.50\) years and \(13.75\) years; errors: about \(0.51\) year and \(0.52\) year
Half-life and doubling time models 52
A pain medication has a half-life of \(3\) hours in the human body. At time \(t=0\), Elena has \(600\,\text{mg}\) of the medication in the bloodstream. a) Find the amount remaining after \(6\), \(9\), and \(12\) hours. b) After how many hours will the amount decrease to \(5\,\text{mg}\)? Round to the nearest hundredth of an hour. c) What percent of the initial amount remains after \(10\) hours? Round to the nearest hundredth of a percent.

Hints

- Count how many half-life intervals fit into each given time. - Write a general exponential decay model using the half-life. - The half-life determines the exponent \(\frac{t}{3}\). - Use logarithms when solving for a time in the exponent. - To find a percent of the initial amount, use the remaining fraction from the model.

Solution

1. The decay model is \(N(t)=600\cdot0.5^{t/3}\). 2. \(N(6)=600\cdot0.5^2=150\,\text{mg}\), \(N(9)=600\cdot0.5^3=75\,\text{mg}\), and \(N(12)=600\cdot0.5^4=37.5\,\text{mg}\). 3. Solve \(600\cdot0.5^{t/3}=5\). Then \(0.5^{t/3}=\frac{1}{120}\). 4. Thus, \(t=3\cdot\frac{\ln(1/120)}{\ln(0.5)}\approx20.72\) hours. 5. The remaining fraction after \(10\) hours is \(0.5^{10/3}\approx0.0992\), or about \(9.92\%\).

Answer

a) After \(6\) hours: \(150\,\text{mg}\); after \(9\) hours: \(75\,\text{mg}\); after \(12\) hours: \(37.5\,\text{mg}\) b) About \(20.72\) hours c) About \(9.92\%\)
Half-life and doubling time models 53
Radon-222 has a half-life of \(3.8\) days. a) What percent of an initial amount remains after \(14\) days? Round to the nearest hundredth of a percent. b) After how many days will the amount decrease to \(10\%\) of its initial value? Round to the nearest hundredth of a day. c) After how many half-lives, and after how many days, will only \(\frac{1}{64}\) of the original amount remain?

Hints

- Use \(100\%\) or \(1\) as the initial amount. - Express elapsed time as a number of half-life intervals. - Relate fractions such as \(\frac{1}{2}\), \(\frac{1}{4}\), and \(\frac{1}{8}\) to powers of \(\frac{1}{2}\). - Multiply the number of half-lives by \(3.8\) days.

Solution

1. Model the percent remaining by \(N(t)=100\cdot0.5^{t/3.8}\). 2. \(N(14)=100\cdot0.5^{14/3.8}\approx7.78\%\). 3. Solve \(0.5^{t/3.8}=0.10\). Then \(t=3.8\cdot\frac{\ln(0.1)}{\ln(0.5)}\approx12.62\) days. 4. Since \(\frac{1}{64}=\left(\frac{1}{2}\right)^6\), the amount reaches this level after \(6\) half-lives. 5. The corresponding time is \(6\cdot3.8=22.8\) days.

Answer

a) About \(7.78\%\) b) About \(12.62\) days c) \(6\) half-lives, or \(22.8\) days
Half-life and doubling time models 54
A pollutant in a treatment tank decays exponentially. Measurements show that its concentration decreases by \(12\%\) each hour. a) Find the hourly decay factor \(a\). b) What percent of the original amount remains after \(5\) hours? Round to the nearest tenth of a percent. c) Find the half-life \(T_{1/2}\) of the pollutant concentration. Round to the nearest hundredth of an hour. d) What hourly percent decrease \(p\) would produce a half-life of exactly \(4\) hours? Round to the nearest tenth of a percent.

Hints

- Convert the percent decrease to a multiplier. - Raise the hourly factor to the number of hours. - At the half-life, \(50\%\) remains. - In part d, find the factor whose fourth power is \(0.5\).

Solution

1. The hourly decay factor is \(a=1-0.12=0.88\). 2. After \(5\) hours, the remaining fraction is \(0.88^5\approx0.5277\), or about \(52.8\%\). 3. Solve \(0.88^{T_{1/2}}=0.5\): \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.88)}\approx5.42\) hours. 4. A \(4\)-hour half-life requires \(a^4=0.5\), so \(a=0.5^{1/4}\approx0.8409\). 5. Therefore, \(p=(1-0.8409)\cdot100\%\approx15.9\%\).

Answer

a) \(a=0.88\) b) About \(52.8\%\) c) \(T_{1/2}\approx5.42\,\text{hours}\) d) About \(15.9\%\) per hour
Half-life and doubling time models 55
A window film reduces the intensity of transmitted light by \(12\%\) for each millimeter of thickness. 1. What percent of the original light intensity \(I_0\) remains after \(3\,\text{mm}\)? Round to the nearest hundredth of a percent. 2. Find the thickness, to the nearest hundredth of a millimeter, that reduces light intensity to one-half of its original value. 3. How thick must the film be so that only \(10\%\) of the original intensity is transmitted? Round to the nearest hundredth of a millimeter.

Hints

- Convert the percent decrease to a multiplicative factor. - Raise the factor to the number of millimeters. - Use logarithms to solve for an unknown exponent. - Write each target as an equation of the form \(q^x=y\).

Solution

1. The transmitted fraction is multiplied by \(0.88\) each millimeter. 2. After \(3\,\text{mm}\), the remaining fraction is \(0.88^3=0.681472\), or about \(68.15\%\). 3. For half the original intensity, solve \(0.88^x=0.5\): \(x=\frac{\ln(0.5)}{\ln(0.88)}\approx5.42\,\text{mm}\). 4. For \(10\%\) transmission, solve \(0.88^x=0.1\): \(x=\frac{\ln(0.1)}{\ln(0.88)}\approx18.01\,\text{mm}\).

Answer

1. About \(68.15\%\) 2. About \(5.42\,\text{mm}\) 3. About \(18.01\,\text{mm}\)
Half-life and doubling time models 56
In a lake, light intensity decreases by \(22\%\) of its current value for each meter of depth. Take the surface intensity as \(100\%\). a) Write a model \(I(d)\) for the percent of surface light remaining at depth \(d\) meters. b) What percent of the light at \(2\,\text{m}\) remains at \(5\,\text{m}\)? Round to the nearest hundredth of a percent. c) A diver thinks light takes less extra depth to halve when it is already weak. Is that correct in this model? Find the extra depth needed to halve the intensity, starting at either \(2\,\text{m}\) or \(5\,\text{m}\). Round to the nearest hundredth of a meter.

Hints

- What fraction of the light remains after each additional meter? - For part b, do you need to know the light at the surface to compare two depths? - For part c, how could you test the diver's claim at both starting depths?

Solution

1. Each meter leaves \(78\%\) of the light from the preceding meter. Thus, \(I(d)=100(0.78)^d\), where \(I(d)\) is a percentage. 2. The ratio of the values at \(5\,\text{m}\) and \(2\,\text{m}\) is \(\frac{I(5)}{I(2)}=0.78^{5-2}=0.78^3=0.474552\). So about \(47.46\%\) of the light at \(2\,\text{m}\) remains at \(5\,\text{m}\). 3. For any starting depth \(d\), the ratio after another \(h\) meters is \(\frac{I(d+h)}{I(d)}=0.78^h\). The starting depth cancels. 4. To halve the intensity, solve \(0.78^h=0.5\). Thus \(h=\frac{\ln(0.5)}{\ln(0.78)}\approx2.79\,\text{m}\), regardless of whether the starting depth is \(2\) or \(5\) meters. The diver's claim is false.

Answer

a) \(I(d)=100(0.78)^d\) b) About \(47.46\%\) c) No. In both cases it takes about \(2.79\,\text{m}\) more depth for the light to halve.
Half-life and doubling time models 57
A radioactive sample has a half-life of \(5\) days. Its initial mass is \(640\,\text{mg}\). a) Find the mass remaining after \(2\), \(3\), and \(5\) half-life intervals. Make a table and give each mass in both milligrams and grams. b) Estimate the mass remaining after \(10\) half-life intervals, or \(50\) days. Use \(\left(\frac{1}{2}\right)^{10}\approx\frac{1}{1000}\).

Hints

- Halving during each interval means repeatedly multiplying by the same factor. - Convert milligrams to grams by dividing by \(1000\). - Write a formula for the mass after \(n\) half-life intervals. - For the estimate, use the approximation supplied in the problem.

Solution

1. The mass is multiplied by \(0.5\) during each half-life interval, so \(M(n)=640(0.5)^n\) milligrams. 2. After \(2\) intervals, \(M(2)=640\cdot(0.5)^2=160\,\text{mg}=0.16\,\text{g}\). 3. After \(3\) intervals, \(M(3)=640\cdot(0.5)^3=80\,\text{mg}=0.08\,\text{g}\). 4. After \(5\) intervals, \(M(5)=640\cdot(0.5)^5=20\,\text{mg}=0.02\,\text{g}\). 5. Using the given estimate, \(M(10)\approx640\cdot\frac{1}{1000}=0.64\,\text{mg}\).

Answer

a) <table> <tr> <th>Half-life intervals \(n\)</th> <th>Mass in mg</th> <th>Mass in g</th> </tr> <tr> <td>2</td> <td>\(160\)</td> <td>\(0.16\)</td> </tr> <tr> <td>3</td> <td>\(80\)</td> <td>\(0.08\)</td> </tr> <tr> <td>5</td> <td>\(20\)</td> <td>\(0.02\)</td> </tr> </table> b) About \(0.64\,\text{mg}\)
Half-life and doubling time models 58
A laboratory begins with \(80\,\text{mg}\) of iodine-131. After exactly \(16\) days, \(20\,\text{mg}\) remains. a) Write a function of the form \(m(t)=m_0a^t\) that gives the mass \(m\), in milligrams, after \(t\) days. b) Find the mass remaining after \(24\) days.

Hints

- Identify the value at \(t=0\). - Use the known mass after \(16\) days to solve for the daily factor. - Notice how many halvings reduce \(80\) to \(20\). - Substitute \(t=24\) into the completed model.

Solution

1. The initial value is \(m_0=80\). 2. Use the point \((16,20)\): \(80a^{16}=20\), so \(a^{16}=0.25\) and \(a=0.25^{1/16}=0.5^{1/8}\approx0.9170\). 3. An exact model is \(m(t)=80(0.5)^{t/8}\). In the requested form, this is \(m(t)=80\left(0.5^{1/8}\right)^t\), or approximately \(m(t)=80(0.9170)^t\). 4. \(m(24)=80\cdot(0.5)^{24/8}=80\cdot(0.5)^3=10\,\text{mg}\).

Answer

a) \(m(t)=80\left(0.5^{1/8}\right)^t\), or approximately \(m(t)=80(0.9170)^t\) b) \(10\,\text{mg}\)
Half-life and doubling time models 59
A forest currently contains \(25{,}000\,\text{m}^3\) of timber and grows by \(2.8\%\) each year. a) Find the timber volume after \(10\) years and after \(20\) years. Round each volume to the nearest cubic meter. b) Find the doubling time \(T_d\). Round to the nearest hundredth of a year. c) After how many years will the original timber volume have tripled? Round to the nearest hundredth of a year.

Hints

- Convert the annual percent increase to a growth factor. - Doubling means the model is twice its initial value. - The initial value cancels when finding a growth-multiple time. - Use logarithms to solve for the time in the exponent.

Solution

1. The growth factor is \(1.028\), so \(W(t)=25{,}000\cdot1.028^t\). 2. \(W(10)\approx32{,}951.19\,\text{m}^3\) and \(W(20)\approx43{,}431.25\,\text{m}^3\), which round to \(32{,}951\,\text{m}^3\) and \(43{,}431\,\text{m}^3\). 3. For doubling, solve \(1.028^{T_d}=2\): \(T_d=\frac{\ln2}{\ln(1.028)}\approx25.10\) years. 4. For tripling, solve \(1.028^t=3\): \(t=\frac{\ln3}{\ln(1.028)}\approx39.78\) years.

Answer

a) After \(10\) years: about \(32{,}951\,\text{m}^3\); after \(20\) years: about \(43{,}431\,\text{m}^3\) b) \(T_d\approx25.10\,\text{years}\) c) About \(39.78\) years
Half-life and doubling time models 60
Marina takes a tablet containing \(400\,\text{mg}\) of medication. The amount in the bloodstream decreases by \(15\%\) each hour. a) Create a table showing the modeled amount remaining after \(0\), \(1\), \(2\), and \(3\) hours. b) Find the medication's half-life \(T_{1/2}\). Round to the nearest hundredth of an hour. c) For this mathematical model, suppose the amount is called “mostly eliminated” when less than \(50\,\text{mg}\) remains. After how many complete hours does this first occur?

Hints

- What fraction of the medication remains each hour? - How can you fill the table with repeated multiplication? - What fraction remains after one half-life? - Compare whole-hour values around the threshold from part c).

Solution

1. The decay model is \(N(t)=400(0.85)^t\). 2. \(N(0)=400\), \(N(1)=340\), \(N(2)=289\), and \(N(3)=245.65\), all in milligrams. 3. For the half-life, solve \(0.85^{T_{1/2}}=0.5\): \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.85)}\approx4.27\) hours. 4. The boundary satisfies \(400(0.85)^t=50\), or \(0.85^t=0.125\), giving \(t\approx12.80\) hours. 5. After \(12\) hours, about \(56.90\,\text{mg}\) remains; after \(13\) hours, about \(48.36\,\text{mg}\) remains. The first whole-hour checkpoint below the set threshold is \(13\) hours.

Answer

a) <table> <tr><td>Time (hours)</td><td>0</td><td>1</td><td>2</td><td>3</td></tr> <tr><td>Amount (mg)</td><td>400</td><td>340</td><td>289</td><td>245.65</td></tr> </table> b) \(T_{1/2}\approx4.27\,\text{hours}\) c) After \(13\) complete hours
Half-life and doubling time models 61
An algae population in a pond grows by \(18\%\) per day under ideal conditions. At the start of the observation, the algae cover \(2500\,\text{cm}^2\). a) Find the doubling time \(T_d\) of the covered area. Round to the nearest hundredth of a day. b) What area is covered after \(12\) days? Round to the nearest square centimeter. c) After how many days will the covered area exceed \(2\,\text{m}^2\)? Round the boundary time to the nearest tenth of a day.

Hints

- Convert the daily percent increase to a growth factor. - Doubling means the area is twice its initial value. - Convert square meters to square centimeters before comparing areas. - Use logarithms to solve for time.

Solution

1. The growth factor is \(1.18\), so \(A(t)=2500\cdot1.18^t\), where \(A\) is measured in \(\text{cm}^2\). 2. For doubling, solve \(1.18^{T_d}=2\): \(T_d=\frac{\ln2}{\ln(1.18)}\approx4.19\) days. 3. \(A(12)=2500\cdot1.18^{12}\approx18{,}218.98\,\text{cm}^2\), or about \(18{,}219\,\text{cm}^2\). 4. Since \(2\,\text{m}^2=20{,}000\,\text{cm}^2\), solve \(2500\cdot1.18^t=20{,}000\). Then \(1.18^t=8\), so \(t=\frac{\ln8}{\ln(1.18)}\approx12.56\) days. 5. The boundary time rounds to \(12.6\) days; the area exceeds \(2\,\text{m}^2\) for times after that boundary.

Answer

a) \(T_d\approx4.19\,\text{days}\) b) About \(18{,}219\,\text{cm}^2\) c) Boundary time: about \(12.6\) days
Half-life and doubling time models 62
Cesium-137, \(^{137}\text{Cs}\), has a half-life of \(30.17\) years. a) Find the annual decay factor \(b\) and the annual percent decrease. Round the factor to four decimal places and the percent to the nearest hundredth. b) A soil sample has an initial activity of \(600\,\text{Bq}\). Find its activity after \(50\) years. Round to the nearest hundredth of a becquerel.

Hints

- A half-life gives the total decay factor over \(30.17\) years. - Relate the annual decay factor to the annual percent decrease. - Use an exponential decay model. - Express \(50\) years as a fraction of one half-life.

Solution

1. The annual decay factor satisfies \(b^{30.17}=0.5\), so \(b=0.5^{1/30.17}\approx0.977287\). 2. The annual percent decrease is \((1-b)\cdot100\%\approx2.27\%\). 3. The activity after \(50\) years is \(A(50)=600\cdot(0.5)^{50/30.17}\approx190.22\,\text{Bq}\).

Answer

a) \(b\approx0.9773\); annual decrease about \(2.27\%\) b) About \(190.22\,\text{Bq}\)
Half-life and doubling time models 63
Iodine-131 (\(^{131}\text{I}\)) has a half-life of \(8.02\) days. a) Find its daily percent decay rate. Round to the nearest hundredth of a percent. b) After how many days will only \(10\%\) of the original amount remain? Round to the nearest tenth of a day.

Hints

- Relate the daily decay factor to what happens after one half-life. - Express \(10\%\) as a decimal fraction of the initial amount. - Use logarithms to solve for an unknown time. - You may write the model as \(N(t)=N_0b^t\).

Solution

1. If \(b\) is the daily decay factor, then \(b^{8.02}=0.5\). 2. Thus, \(b=0.5^{1/8.02}\approx0.9172\). 3. The daily percent decay rate is \((1-0.9172)\cdot100\%\approx8.28\%\). 4. Solve \(0.5^{t/8.02}=0.1\). 5. Therefore, \(t=8.02\cdot\frac{\log(0.1)}{\log(0.5)}\approx26.64\) days, which rounds to \(26.6\) days.

Answer

a) About \(8.28\%\) per day b) About \(26.6\) days
Half-life and doubling time models 64
A laboratory studies a \(400\,\text{mg}\) sample of strontium-90 that decreases by \(2.4\%\) each year. a) Find the mass after \(15\) years. Round to the nearest hundredth of a milligram. b) By what percent does the mass decrease over those \(15\) years? Round to the nearest hundredth of a percent. c) Find the half-life \(T_{1/2}\). Round to the nearest tenth of a year.

Hints

- Convert the annual percent decrease to a decay factor. - Write an exponential decay model. - Compare the remaining fraction with \(1\) to find the percent decrease. - At the half-life, one half of the original mass remains. - Use logarithms to solve for time.

Solution

1. The yearly decay factor is \(0.976\), so \(N(t)=400\cdot0.976^t\). 2. \(N(15)=400\cdot0.976^{15}\approx277.85\,\text{mg}\). 3. The fraction lost is \(1-0.976^{15}\approx0.30538\), so the total decrease is about \(30.54\%\). 4. Solve \(0.976^{T_{1/2}}=0.5\): \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.976)}\approx28.53\) years, which rounds to \(28.5\) years.

Answer

a) About \(277.85\,\text{mg}\) b) About \(30.54\%\) c) \(T_{1/2}\approx28.5\,\text{years}\)
Half-life and doubling time models 65
Carbon-14 has a half-life of \(5730\) years and can be used to date archaeological samples. In the same measurement setup, an excavated bone sample has carbon-14 activity of \(3.5\) relative units for every \(12.0\) relative units measured in comparable modern material. a) Find the annual decay factor \(b\). Round to six decimal places. b) Estimate the age of the bone sample to the nearest year.

Hints

- Relate the annual decay factor to the half-life. - Compare the sample's activity with the modern-material activity as a ratio. - Use logarithms to isolate the time exponent. - Keep enough decimal places in the decay factor, or use its exact expression, to avoid a large rounding error.

Solution

1. The annual decay factor satisfies \(b^{5730}=0.5\). 2. Therefore, \(b=0.5^{1/5730}\approx0.999879\). 3. The activity ratio is \(\frac{3.5}{12.0}\), so \(\frac{3.5}{12.0}=b^t\). 4. To avoid rounding error, substitute \(b=0.5^{1/5730}\): \(t=5730\cdot\frac{\ln(3.5/12.0)}{\ln(0.5)}\approx10{,}185.69\). 5. The bone sample is about \(10{,}186\) years old.

Answer

a) \(b\approx0.999879\) b) About \(10{,}186\) years old
Half-life and doubling time models 66
A radioactive isotope used in medical imaging has an activity of \(400\,\text{MBq}\) at the time of injection (\(t=0\)). After exactly \(4\) hours, its activity is \(240\,\text{MBq}\). Assume an exponential decay model \(A(t)=A_0e^{kt}\), where \(t\) is measured in hours. a) Find \(k\), rounded to four decimal places. b) Find the half-life \(T_{1/2}\), rounded to the nearest hundredth of an hour. c) Find when the modeled activity falls to \(5\%\) of its initial value. Round to the nearest hundredth of an hour.

Hints

- What fraction of the original activity remains after four hours? - How can that fraction help you find the exponent coefficient? - What fraction remains after one half-life? - For the stipulated cutoff, what ratio of remaining to initial activity is needed?

Solution

1. Using \(A(4)=240\), \(240=400e^{4k}\), so \(e^{4k}=0.6\). 2. Therefore, \(k=\frac{\ln(0.6)}{4}\approx-0.1277\,\text{h}^{-1}\). 3. For the half-life, solve \(e^{kT_{1/2}}=0.5\). Thus, \(T_{1/2}=\frac{\ln(0.5)}{k}\approx5.43\) hours. 4. For \(5\%\) remaining, solve \(e^{kt}=0.05\). Thus, \(t=\frac{\ln(0.05)}{k}\approx23.46\) hours.

Answer

a) \(k\approx-0.1277\,\text{h}^{-1}\) b) \(T_{1/2}\approx5.43\,\text{hours}\) c) About \(23.46\) hours
Half-life and doubling time models 67
A medical contrast agent has a half-life of \(3\) hours in Kenji's bloodstream. Kenji receives \(120\,\text{mg}\) of the agent. 1. Write a function \(M(t)\) for the amount remaining after \(t\) hours. 2. Find the amount remaining after \(10\) hours. Round to the nearest hundredth of a milligram. 3. Find when only \(10\%\) of the initial amount remains. Round to the nearest hundredth of an hour.

Hints

- Express the number of half-life intervals as a ratio involving \(t\). - For part 2, substitute the given time into the half-life model. - Ten percent remaining means the amount is \(0.10\) times the initial amount. - Use logarithms to solve for the unknown time in part 3.

Solution

1. With a half-life of \(3\) hours, the model is \(M(t)=120(0.5)^{t/3}\). 2. \(M(10)=120(0.5)^{10/3}\approx11.91\,\text{mg}\). 3. Set \((0.5)^{t/3}=0.10\). Taking logarithms gives \(t=\frac{3\ln(0.10)}{\ln(0.5)}\approx9.97\) hours.

Answer

1. \(M(t)=120(0.5)^{t/3}\) 2. About \(11.91\,\text{mg}\) 3. About \(9.97\,\text{hours}\)
Half-life and doubling time models 68
An invasive aquatic plant spreads across a lake. a) The covered area doubles every \(5\) days and is initially \(1\,\text{m}^2\). Write a function \(A(t)\) for the area after \(t\) days. b) Use \(2^{10}\approx10^3\) to estimate the area predicted by the unlimited model after \(50\) days and after \(100\) days. c) The lake has an area of \(1\,\text{km}^2\). Estimate when half the lake, or \(500{,}000\,\text{m}^2\), is covered. Use \(500\approx2^9\). d) Use logarithms to find an exact expression for the model time when half the lake is covered, then approximate it to the nearest hundredth of a day and compare it with the estimate from part c).

Hints

- Express the number of doubling intervals using \(t\) and the doubling time. - Convert the lake area to square meters before comparing areas. - For the estimate, combine the two supplied powers of \(2\). - For the exact model time, keep the target value inside a logarithm instead of replacing it by a nearby power of \(2\).

Solution

1. Since the area doubles every \(5\) days, \(A(t)=2^{t/5}\). 2. \(A(50)=2^{10}\approx1000\,\text{m}^2\). Also, \(A(100)=2^{20}=(2^{10})^2\approx1{,}000{,}000\,\text{m}^2\). The second value is only a prediction of the unlimited model. 3. \(500{,}000=500\cdot1000\approx2^9\cdot2^{10}=2^{19}\). Thus, \(t/5\approx19\), so \(t\approx95\) days. 4. Solve \(2^{t/5}=500000\): \(t=\frac{5\ln(500000)}{\ln 2}\approx94.66\) days. 5. The estimate from part c) is about \(95-94.66=0.34\) day greater.

Answer

a) \(A(t)=2^{t/5}\) b) \(A(50)\approx1000\,\text{m}^2\); \(A(100)\approx1{,}000{,}000\,\text{m}^2\) c) About \(95\,\text{days}\) d) \(t=\frac{5\ln(500000)}{\ln 2}\approx94.66\,\text{days}\); the estimate is about \(0.34\) day greater.
Half-life and doubling time models 69
A bacterial culture doubles every \(3\) hours. Its population is modeled by \(N(t)=N_0b^t\), where \(t\) is measured in hours. a) Find \(b\) exactly and rounded to four decimal places. b) By what factor does the modeled population change when you move \(1\) hour backward in time? Give the exact factor and its value to four decimal places. c) By what factor does it change over \(4.5\) hours? Give the exact factor and its value to four decimal places. d) Find the percent increase during \(30\) minutes. Round to the nearest hundredth of a percent.

Hints

- How does doubling over three hours constrain the hourly factor? - What happens to the exponent when time moves backward? - A time interval need not be a whole number of hours. - Compare a change factor with \(1\) to express the change as a percent.

Solution

1. The doubling condition gives \(b^3=2\), so \(b=2^{1/3}\approx1.2599\). 2. Moving one hour backward divides the modeled population by \(b\), giving factor \(b^{-1}=2^{-1/3}\approx0.7937\). 3. The \(4.5\)-hour factor is \(b^{4.5}=2^{4.5/3}=2^{3/2}=\sqrt{8}\approx2.8284\). 4. Thirty minutes is \(0.5\) hour. The factor is \(b^{0.5}=2^{1/6}\), so the percent increase is \((2^{1/6}-1)\cdot100\%\approx12.25\%\).

Answer

a) \(b=2^{1/3}\approx1.2599\) b) \(2^{-1/3}\approx0.7937\) c) \(\sqrt{8}\approx2.8284\) d) About \(12.25\%\)
Half-life and doubling time models 70
The concentration of a medication in the blood is modeled by an exponential function. The table gives the concentration \(y\), in \(\text{mg/L}\), after \(t\) hours. <table><tr><td>\(t\)</td><td>0</td><td>2</td><td>5</td></tr><tr><td>\(y\)</td><td>\(12.0\)</td><td>\(7.68\)</td><td>\(3.93216\)</td></tr></table> a) Find a function of the form \(f(t)=ca^t\). b) Find the percent decrease per hour. c) Find the half-life \(T_{1/2}\), rounded to the nearest hundredth of an hour.

Hints

- Use the value at \(t=0\) to determine the initial-value parameter. - Use another table value to determine the hourly decay factor, then check it against the remaining table value. - Interpret the hourly factor as the fraction that remains after one hour. - At the half-life, one half of the initial concentration remains.

Solution

1. Since \(f(0)=12.0\), \(c=12\). Using \(f(2)=7.68\), \(12a^2=7.68\), so \(a^2=0.64\) and \(a=0.8\). Thus, \(f(t)=12(0.8)^t\). The value at \(t=5\) confirms the model. 2. The factor \(0.8\) means \(80\%\) remains each hour, so the concentration decreases by \(20\%\) per hour. 3. For the half-life, solve \((0.8)^{T_{1/2}}=0.5\). Therefore, \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.8)}\approx3.11\) hours.

Answer

a) \(f(t)=12(0.8)^t\) b) \(20\%\) per hour c) \(T_{1/2}\approx3.11\,\text{hours}\)
Half-life and doubling time models 71
The concentration of a medication in the blood is modeled by \(C(t)=C_0e^{-kt}\), where \(C_0=250\,\text{mg/L}\), \(k=0.085\,\text{h}^{-1}\), and \(t\) is measured in hours. a) Find when the concentration decreases to \(100\,\text{mg/L}\). Round to the nearest hundredth of an hour. b) Find the half-life \(T_{1/2}\). Round to the nearest hundredth of an hour. c) Find the percent decrease per hour, rounded to the nearest hundredth of a percent. Compare this percent with the numerical value of \(k\).

Hints

- Compare the target concentration with the initial concentration before solving for time. - At the half-life, the concentration ratio is one half. - The one-hour remaining factor comes from evaluating the exponential factor over one hour. - Distinguish the continuous decay constant from a discrete one-hour percent change.

Solution

1. Set \(C(t)=100\): \(100=250e^{-0.085t}\). Thus, \(e^{-0.085t}=0.4\), so \(t=\frac{\ln(0.4)}{-0.085}\approx10.78\) hours. 2. The half-life satisfies \(e^{-0.085T_{1/2}}=0.5\). Therefore, \(T_{1/2}=\frac{\ln(0.5)}{-0.085}\approx8.15\) hours. 3. The one-hour factor is \(e^{-0.085}\approx0.9185\). Therefore, the hourly decrease is \((1-e^{-0.085})\cdot100\%\approx8.15\%\). 4. This is slightly less than \(8.5\%\); \(k=0.085\,\text{h}^{-1}\) is a continuous decay constant, not the discrete hourly percent decrease.

Answer

a) About \(10.78\,\text{hours}\) b) \(T_{1/2}\approx8.15\,\text{hours}\) c) About \(8.15\%\) per hour, which is slightly less than \(8.5\%\)
Half-life and doubling time models 72
A wetland had \(1200\) sandhill cranes in 2015. Their number is modeled by \(B(t)=B_0e^{a(t-t_0)}\), with continuous growth rate \(a=0.045\,\text{yr}^{-1}\), where \(t\) is the calendar year and \(t_0=2015\). a) Find the predicted number of cranes in 2030. Round to the nearest whole crane. b) Determine during which year the model reaches \(2400\) cranes. c) Find the effective annual percent growth rate, rounded to the nearest hundredth of a percent, and explain why it is greater than \(4.5\%\).

Hints

- What elapsed time separates 2015 from 2030? - What factor represents doubling from the initial population? - How does the exponent in the given model change from one calendar year to the next? - Compare that one-year factor with the factor for a \(4.5\%\) increase.

Solution

1. From 2015 to 2030 is \(15\) years. Thus, \(B(2030)=1200e^{0.045(15)}\approx2356.84\), which rounds to \(2357\) cranes. 2. To double, solve \(e^{0.045\Delta t}=2\). Then \(\Delta t=\frac{\ln(2)}{0.045}\approx15.40\) years, so the population reaches \(2400\) during 2030. 3. The one-year factor is \(e^{0.045}\approx1.0460\). The effective annual growth rate is \((e^{0.045}-1)\cdot100\%\approx4.60\%\). 4. A \(4.5\%\) annual increase would have a one-year factor of \(1.045\), which is smaller than \(e^{0.045}\). That is why the effective annual increase exceeds \(4.5\%\).

Answer

a) About \(2357\) cranes b) During 2030 c) About \(4.60\%\) per year, because the one-year factor \(e^{0.045}\) is greater than \(1.045\).
Half-life and doubling time models 73
Tritium is a radioactive isotope of hydrogen with a half-life of \(12.32\) years. Because small amounts occur in rainwater, tritium can be used to estimate the age of wine. a) Write a model of the form \(N(t)=N_0a^t\) for the amount remaining after \(t\) years. Round \(a\) to four decimal places. b) Find the annual percent decrease in tritium. Round to the nearest hundredth of a percent. c) A wine sample contains \(22\%\) of its original tritium. Estimate the wine's age to the nearest year.

Hints

- Use the half-life to determine the one-year remaining factor. - Compare the remaining factor with \(1\) to find the annual percent decrease. - Write \(22\%\) as a decimal fraction of the original amount. - Use logarithms when solving for the age.

Solution

1. The half-life gives \(a^{12.32}=0.5\). Therefore, \(a=(0.5)^{1/12.32}\approx0.9453\), so \(N(t)=N_0(0.9453)^t\). 2. The annual decrease is \((1-0.9453)\cdot100\%\approx5.47\%\). 3. Solve \((0.9453)^t=0.22\). Then \(t=\frac{\ln(0.22)}{\ln(0.9453)}\approx26.91\), so the wine is about \(27\) years old.

Answer

a) \(N(t)=N_0(0.9453)^t\) b) About \(5.47\%\) per year c) About \(27\,\text{years}\)
Half-life and doubling time models 74
A radioactive substance decays exponentially. After \(12\) days, exactly \(75\%\) of the initial mass remains. 1. Find the daily decay factor \(q\), rounded to four decimal places, and the decay constant \(\lambda\) for \(M(t)=M_0e^{-\lambda t}\), rounded to five decimal places, where \(t\) is measured in days. 2. Find the half-life \(T_{1/2}\), rounded to the nearest hundredth of a day. 3. Find when only \(10\%\) of the initial mass remains. Round to the nearest hundredth of a day.

Hints

- Use the twelve-day remaining fraction to determine the one-day factor. - Relate the discrete daily factor to the base-\(e\) form. - At the half-life, the remaining fraction is one half. - For the final part, compare the remaining amount with the initial amount as a ratio.

Solution

1. Since \(q^{12}=0.75\), \(q=(0.75)^{1/12}\approx0.9763\). Also, \(e^{-12\lambda}=0.75\), so \(\lambda=-\frac{\ln(0.75)}{12}\approx0.02397\,\text{day}^{-1}\). 2. The half-life is \(T_{1/2}=\frac{\ln 2}{\lambda}\approx28.91\) days. 3. For \(10\%\) remaining, solve \(e^{-\lambda t}=0.10\). Thus, \(t=\frac{-\ln(0.10)}{\lambda}\approx96.05\) days.

Answer

1. \(q\approx0.9763\); \(\lambda\approx0.02397\,\text{day}^{-1}\) 2. \(T_{1/2}\approx28.91\,\text{days}\) 3. About \(96.05\,\text{days}\)
Half-life and doubling time models 75
Two bacterial cultures are being studied in a laboratory. Culture A doubles every \(5\) hours. Culture B doubles every \(8\) hours but begins with four times as many bacteria as Culture A. Find when the two cultures will have the same number of bacteria.

Hints

- Write an exponential model for each culture. - Use a variable for Culture A's unknown initial population. - Rewrite \(4\) as a power of \(2\). - Once both sides have the same base, compare the exponents.

Solution

1. Let Culture A begin with \(N_0\) bacteria. Then \(N_A(t)=N_0 2^{t/5}\) and \(N_B(t)=4N_0 2^{t/8}\). 2. Set the populations equal: \(N_0 2^{t/5}=4N_0 2^{t/8}\). Dividing by \(N_0\) gives \(2^{t/5}=4\cdot 2^{t/8}\). 3. Since \(4=2^2\), \(2^{t/5}=2^{2+t/8}\), so \(\frac{t}{5}=2+\frac{t}{8}\). 4. Solving gives \(\frac{3t}{40}=2\), so \(t=\frac{80}{3}=26\frac{2}{3}\) hours.

Answer

The cultures have the same population after \(26\frac{2}{3}\) hours, or about \(26.67\) hours.
Half-life and doubling time models 76
Two animal populations are being monitored in a wildlife reserve. At the start of the study, Species A has about \(15{,}000\) animals and grows by \(2.1\%\) each year. Species B has \(8000\) animals and grows by \(4.5\%\) each year. a) Find the doubling time \(T_d\) for Species B. Round to the nearest hundredth of a year. b) After how many years will the population of Species A be only \(1.5\) times the population of Species B? Round to the nearest hundredth of a year.

Hints

- Convert each percent increase to a yearly growth factor. - A doubling time makes the final-to-initial ratio equal to \(2\). - For part b), translate “\(1.5\) times” into an equation relating the two models. - Use logarithms to solve an equation with the variable in an exponent.

Solution

1. The population models are \(A(t)=15{,}000(1.021)^t\) and \(B(t)=8000(1.045)^t\). 2. For Species B to double, solve \((1.045)^{T_d}=2\). Thus, \(T_d=\frac{\ln 2}{\ln(1.045)}\approx15.75\) years. 3. For part b), set \(A(t)=1.5B(t)\): \(15{,}000(1.021)^t=1.5\cdot8000(1.045)^t\). 4. This simplifies to \(1.25=\left(\frac{1.045}{1.021}\right)^t\), so \(t=\frac{\ln(1.25)}{\ln(1.045)-\ln(1.021)}\approx9.60\) years.

Answer

a) \(T_d\approx15.75\,\text{years}\) b) About \(9.60\,\text{years}\)
Half-life and doubling time models 77
A technology company tracks registered users of a new app. <table style="width: 100%; border-collapse: collapse;"><thead><tr style="border-bottom: 1px solid black;"><th style="text-align: left; padding: 5px;">Year</th><th style="text-align: left; padding: 5px;">Registered users</th></tr></thead><tbody><tr><td style="padding: 5px;">2016</td><td style="padding: 5px;">\(120{,}000\)</td></tr><tr><td style="padding: 5px;">2018</td><td style="padding: 5px;">\(240{,}000\)</td></tr><tr><td style="padding: 5px;">2021</td><td style="padding: 5px;">\(480{,}000\)</td></tr><tr><td style="padding: 5px;">2025</td><td style="padding: 5px;">\(960{,}000\)</td></tr></tbody></table> a) Find each doubling interval. b) Find the average annual growth rate during each doubling interval. Round each rate to the nearest hundredth of a percent. c) Interpret how the changing doubling intervals reflect the app's growth dynamics.

Hints

- Compare the years associated with consecutive doubled values. - For a doubling interval \(T_d\), the yearly factor satisfies \(q^{T_d}=2\). - Convert each yearly factor to a percent increase by comparing it with \(1\). - Longer doubling intervals correspond to smaller annual percent growth rates.

Solution

1. The user count doubles from 2016 to 2018 in \(2\) years, from 2018 to 2021 in \(3\) years, and from 2021 to 2025 in \(4\) years. 2. The yearly factors are \(2^{1/2}\approx1.4142\), \(2^{1/3}\approx1.2599\), and \(2^{1/4}\approx1.1892\). 3. The corresponding annual growth rates are about \(41.42\%\), \(25.99\%\), and \(18.92\%\). 4. The doubling intervals are getting longer, so the relative annual growth rate is decreasing even though the total number of users continues to increase.

Answer

a) \(2\) years, \(3\) years, and \(4\) years b) About \(41.42\%\), \(25.99\%\), and \(18.92\%\) per year c) The relative growth rate is decreasing because each doubling takes longer.
Half-life and doubling time models 78
A threatened animal population in a wildlife reserve is assumed to grow exponentially. At the start of the study, there are \(200\) animals. After \(5\) years, there are \(350\). 1. Find the annual growth factor \(q\), rounded to four decimal places, and the annual percent growth rate, rounded to the nearest hundredth of a percent. 2. Find how long it takes the initial population to quadruple. Round to the nearest hundredth of a year. 3. An equivalent continuous model has the form \(B(t)=B_0e^{kt}\). Find \(k\), rounded to four decimal places, and use the model to predict the population after \(10\) years, rounded to the nearest whole animal.

Hints

- Use the two population values to determine the yearly growth factor. - Quadrupling means the final-to-initial ratio is \(4\). - Use logarithms to solve for an unknown time or continuous growth constant. - Compare the five-year and ten-year growth factors before doing unnecessary computation.

Solution

1. From \(350=200q^5\), \(q=(1.75)^{1/5}\approx1.1184\). The annual growth rate is \((q-1)\cdot100\%\approx11.84\%\). 2. For quadrupling, solve \(q^t=4\). Thus, \(t=\frac{\ln4}{\ln q}\approx12.39\) years. 3. From \(200e^{5k}=350\), \(k=\frac{\ln1.75}{5}\approx0.1119\,\text{yr}^{-1}\). 4. Then \(B(10)=200e^{10k}=200(1.75)^2=612.5\), which rounds to \(613\) animals.

Answer

1. \(q\approx1.1184\); about \(11.84\%\) per year 2. About \(12.39\,\text{years}\) 3. \(k\approx0.1119\,\text{yr}^{-1}\); about \(613\) animals after \(10\) years
Half-life and doubling time models 79
The amount of a medication in Sofia's blood is modeled by \(M(t)=150(0.82)^t\), where \(M(t)\) is measured in milligrams and \(t\) is measured in hours. a) Rewrite the model as \(M(t)=150e^{kt}\). Round \(k\) to four decimal places. b) Find the medication's half-life \(T_{1/2}\), rounded to the nearest hundredth of an hour. c) A second medication starts at \(300\,\text{mg}\) and follows a model \(g(t)=be^{kt}\) with \(k=-0.12\,\text{h}^{-1}\). Write its model and compare its graph with the graph of \(M\).

Hints

- How can you write a factor less than one using a base-e exponent? - At the half-life, what is the remaining fraction? - Compare the starting amounts and the signs and sizes of the exponent coefficients. - What happens to the ratio of the two modeled amounts as time increases?

Solution

1. Since \(e^k=0.82\), \(k=\ln(0.82)\approx-0.1985\,\text{h}^{-1}\). Thus, \(M(t)=150e^{\ln(0.82)t}\approx150e^{-0.1985t}\). 2. Solve \(0.82^{T_{1/2}}=0.5\). Then \(T_{1/2}=\frac{\ln(0.5)}{\ln(0.82)}\approx3.49\) hours. 3. The second model is \(g(t)=300e^{-0.12t}\). It starts at twice the value of \(M\). 4. Its exponent coefficient is less negative, so \(g\) decreases more slowly as a percentage of its current amount. Also, \(\frac{g(t)}{M(t)}=2e^{(-0.12-\ln(0.82))t}>2\) for \(t>0\), so its graph stays above the graph of \(M\).

Answer

a) \(k\approx-0.1985\,\text{h}^{-1}\) b) \(T_{1/2}\approx3.49\) hours c) \(g(t)=300e^{-0.12t}\); it starts higher, remains above \(M\), and decreases more slowly proportionally.
Half-life and doubling time models 80
After Maya takes a medication, the amount of the active ingredient in her body decreases continuously. The graph shows the remaining amount \(W(t)\), in milligrams, after \(t\) hours. a) Use the graph to determine when the initial amount has been reduced to one-half, one-fourth, and one-eighth. What pattern do you notice? b) State the medication's half-life. c) Find the percent decrease per hour. Round to the nearest tenth of a percent. d) Calculate the amount of medication remaining after exactly \(15\) hours.
Figure for problem 532184

Hints

- Read the initial amount from the graph at \(t=0\). - Locate successive fractions of the initial amount on the vertical axis. - Compare the time intervals between successive halvings. - Relate the hourly decay factor to the time required for one halving. - For the last part, count how many half-life intervals fit into \(15\) hours.

Solution

1. The graph starts at \(80\,\text{mg}\). It reaches \(40\,\text{mg}\) after \(3\) hours, \(20\,\text{mg}\) after \(6\) hours, and \(10\,\text{mg}\) after \(9\) hours. Each additional halving takes \(3\) hours. 2. Therefore, \(T_{1/2}=3\) hours. 3. If \(b\) is the hourly decay factor, then \(b^3=0.5\). Thus, \(b=0.5^{1/3}\approx0.7937\), so the hourly percent decrease is \((1-b)\cdot100\%\approx20.6\%\). 4. Fifteen hours is \(5\) half-lives. Therefore, \(W(15)=80(0.5)^5=2.5\,\text{mg}\).

Answer

a) One-half after \(3\) hours, one-fourth after \(6\) hours, and one-eighth after \(9\) hours; each halving takes \(3\) hours. b) \(T_{1/2}=3\,\text{hours}\) c) About \(20.6\%\) per hour d) \(2.5\,\text{mg}\)
Half-life and doubling time models 81
After a dose, the concentration of a medication in Elena's blood follows the graph. a) Use the marked points to find an exponential model of the form \(C(t)=C_0a^t\). Round \(a\) to four decimal places. Also state the medication's half-life \(T_{1/2}\). b) Find when the modeled concentration reaches \(0.1\,\text{mg/L}\). Round to the nearest tenth of an hour. State when the model is below that level.
Figure for problem 532457

Hints

- Read the initial value where the graph meets the y-axis. - How often do the marked concentrations halve? - Convert the characteristic time into a one-hour factor. - How can the ratio of target to initial concentration help you find a time?

Solution

1. The point \((0, 8)\) gives \(C_0=8\). 2. The values halve every \(3\) hours: \(8\) at \(t=0\), \(4\) at \(t=3\), and \(2\) at \(t=6\). Therefore, \(T_{1/2}=3\) hours. 3. Using \((3, 4)\), \(4=8a^3\), so \(a^3=0.5\) and \(a=0.5^{1/3}\approx0.7937\). An exact model is \(C(t)=8(0.5)^{t/3}\). 4. Solve \(8(0.5)^{t/3}=0.1\). Thus, \(t=\frac{3\ln(0.0125)}{\ln(0.5)}\approx18.97\) hours, or \(19.0\) hours to the nearest tenth. Concentration is below that level afterward.

Answer

a) \(C(t)=8(0.5^{1/3})^t\), where \(a\approx0.7937\); \(T_{1/2}=3\) hours b) The level is reached at \(t=\frac{3\ln(0.0125)}{\ln(0.5)}\approx19.0\) hours; it is below \(0.1\,\text{mg/L}\) for times after this exact value.
Half-life and doubling time models 82
The graph shows an exponential growth process \(f\) and an exponential decay process \(g\). Time \(t\) is measured in hours. a) Find the initial amount for each process. b) Use the graph to find the doubling time of \(f\) and the half-life of \(g\). c) Write an equation of the form \(N(t)=N_0b^t\) for each process. d) Find each amount after \(15\) hours. Round the value for \(g\) to the nearest tenth.
Figure for problem 532626

Hints

- The initial amount is the value when \(t=0\). - Locate twice the initial value of \(f\) and one-half the initial value of \(g\). - Use the initial value and the characteristic time to determine each hourly factor. - Substitute \(t=15\) into both equations.

Solution

1. The y-intercepts give \(f(0)=10\) and \(g(0)=80\). 2. The graph shows that \(f\) grows from \(10\) to \(20\) in \(3\) hours, so its doubling time is \(3\) hours. It shows that \(g\) decreases from \(80\) to \(40\) in \(4\) hours, so its half-life is \(4\) hours. 3. For \(f\), the hourly factor is \(2^{1/3}\), so \(f(t)=10\left(2^{1/3}\right)^t=10(2)^{t/3}\). 4. For \(g\), the hourly factor is \(0.5^{1/4}\), so \(g(t)=80\left(0.5^{1/4}\right)^t=80(0.5)^{t/4}\). 5. \(f(15)=10\cdot2^{15/3}=320\), and \(g(15)=80\cdot(0.5)^{15/4}\approx5.9\).

Answer

a) \(f(0)=10\); \(g(0)=80\) b) Doubling time of \(f\): \(3\) hours; half-life of \(g\): \(4\) hours c) \(f(t)=10(2)^{t/3}\); \(g(t)=80(0.5)^{t/4}\) d) \(f(15)=320\); \(g(15)\approx5.9\)
Half-life and doubling time models 83
Iodine-131 is used in some thyroid treatments. a) Determine the half-life \(T_{1/2}\). b) Write the model in the form \(m(t)=m_0\cdot0.5^{t/T_{1/2}}\). c) Calculate the mass remaining after \(20\) days. Round to the nearest hundredth of a milligram.
Figure for problem 534028

Hints

- What initial mass can you read at the graph's y-intercept? - Locate the time when half of that mass remains. - Which values from the graph determine the half-life model? - Substitute the requested time into the model before rounding.

Solution

1. The graph shows an initial mass of \(10\,\text{mg}\), so half of it is \(5\,\text{mg}\). This is reached at \(t=8\) days, giving \(T_{1/2}=8\) days. 2. Substitute \(m_0=10\) and \(T_{1/2}=8\) into the model: \(m(t)=10\cdot0.5^{t/8}\). 3. After \(20\) days, \(m(20)=10\cdot0.5^{20/8}\approx1.77\,\text{mg}\).

Answer

a) \(T_{1/2}=8\) days b) \(m(t)=10\cdot0.5^{t/8}\) c) About \(1.77\,\text{mg}\)
Half-life and doubling time models 84
A biologist observes two bacterial cultures. Culture A starts with \(100\) bacteria and doubles every \(20\) minutes. Culture B starts with \(100\) bacteria and triples every \(30\) minutes. Determine which culture reaches \(1{,}000{,}000\) bacteria first. Find the difference, in minutes, between the two times. Round to the nearest hundredth.

Hints

- Write a growth model for each culture from its repeated multiplication factor. - Which fraction of elapsed time gives the number of doubling or tripling periods? - Find the two times separately, then subtract their full-precision values. - Round only the final difference.

Solution

1. Let \(t\) be time in minutes. The models are \(N_A(t)=100\cdot2^{t/20}\) and \(N_B(t)=100\cdot3^{t/30}\). 2. Culture A reaches the target when \(2^{t/20}=10{,}000\), giving \(t_A=20\frac{\ln(10{,}000)}{\ln(2)}\approx265.75425\) minutes. 3. Culture B reaches the target when \(3^{t/30}=10{,}000\), giving \(t_B=30\frac{\ln(10{,}000)}{\ln(3)}\approx251.50839\) minutes. 4. Culture B is first. Subtract the unrounded times: \(t_A-t_B\approx14.24585\), which rounds to \(14.25\) minutes. Subtracting the times rounded to hundredths would instead give an inaccurate final difference.

Answer

Culture B reaches \(1{,}000{,}000\) bacteria first, about \(14.25\) minutes before Culture A.
Half-life and doubling time models 85
A new social media channel follows the model \(F(t)=S-ce^{-kt}\), where \(t\) is the number of months since it started and \(k>0\). It has \(400\) followers at \(t=0\), \(760\) followers after \(5\) months, and \(976\) followers after \(10\) months. 1. Find the saturation level \(S\), which is the follower count the model approaches over time. 2. Find how many months it takes for the gap between \(F(t)\) and \(S\) to be cut in half. Round to the nearest hundredth of a month.

Hints

- How far is the follower count below its eventual level at each recorded time? - How is the gap after \(10\) months related to the gap after \(5\) months? - Once you know the gap's decay factor over \(5\) months, how could you find the time for that factor to be \(\frac{1}{2}\)?

Solution

1. From \(F(0)=400\), we get \(S-c=400\), so \(c=S-400\). 2. Set \(u=e^{-5k}\). Then \(e^{-10k}=u^2\). The other two measurements give \((S-400)u=S-760\) and \((S-400)u^2=S-976\). 3. Eliminate \(u\): \((S-760)^2=(S-400)(S-976)\). Expanding gives \(S^2-1520S+577{,}600=S^2-1376S+390{,}400\), so \(S=1300\). 4. Then \(c=900\), and \(u=\frac{1300-760}{900}=0.6\). Thus \(e^{-5k}=0.6\), so \(k=-\frac{\ln(0.6)}{5}\). 5. The remaining gap is \(S-F(t)=ce^{-kt}\). It halves after \(T\) months when \(e^{-kT}=\frac{1}{2}\). Hence \(T=\frac{\ln(2)}{k}=\frac{5\ln(2)}{-\ln(0.6)}\approx6.78\) months.

Answer

1. \(S=1300\) followers. 2. About \(6.78\) months.
Half-life and doubling time models 86
Investigate the time \(n\) needed for a quantity growing at an annual rate of \(p\%\) to quadruple. 1. Find the quadrupling time for \(p=2\), \(p=5\), and \(p=10\). Round to the nearest hundredth. 2. For each case, multiply the time \(n\) by the percent rate \(p\). What approximate constant \(c\) do you obtain? Use it to write a rule of thumb in the form \(n\approx\frac{c}{p}\).

Hints

- Write an equation for a quantity that has become four times its initial value. - Use logarithms to find each time. - Examine whether the products \(np\) are approximately constant. - Choose a simple, memorable value for the constant in the rule of thumb.

Solution

1. The quadrupling time is \(n=\frac{\ln(4)}{\ln\left(1+\frac{p}{100}\right)}\). 2. For \(p=2\), \(n\approx70.01\) years and, using the unrounded time, \(np\approx140.01\). 3. For \(p=5\), \(n\approx28.41\) years and, using the unrounded time, \(np\approx142.07\). 4. For \(p=10\), \(n\approx14.55\) years and, using the unrounded time, \(np\approx145.45\). 5. The products are all near \(140\) to \(145\). One simple rule of thumb is \(n\approx\frac{140}{p}\); a constant near \(142\) would also fit these three values.

Answer

1. \(70.01\) years, \(28.41\) years, and \(14.55\) years 2. Using unrounded times, the products \(np\) are about \(140.01\), \(142.07\), and \(145.45\). A reasonable rule is \(n\approx\frac{140}{p}\).
Half-life and doubling time models 87
A radioactive isotope decays so that after \(500\) years, \(25\%\) of the original mass has decayed. a) Find the decay factor \(q\) for a \(500\)-year period. b) Find the annual decay factor \(a\) to four decimal places. c) Find the half-life \(T_{1/2}\) to the nearest year. d) After how many years will only \(1\%\) of the original mass remain? Round to the nearest year.

Hints

- First determine what percent of the mass remains after \(500\) years. - Relate the long-term decay factor to the annual factor using a power. - At the half-life, one half of the initial mass remains. - Use logarithms to solve when the unknown is in the exponent.

Solution

1. After \(500\) years, \(75\%\) remains, so \(q=0.75\). 2. The annual factor satisfies \(a^{500}=0.75\), so \(a=0.75^{1/500}\approx0.9994\). 3. Solve \(0.75^{T_{1/2}/500}=0.5\). Then \(T_{1/2}=500\cdot\frac{\ln(0.5)}{\ln(0.75)}\approx1204.71\), so the half-life is about \(1205\) years. 4. Solve \(0.75^{t/500}=0.01\). Then \(t=500\cdot\frac{\ln(0.01)}{\ln(0.75)}\approx8003.92\), or about \(8004\) years.

Answer

a) \(q=0.75\) b) \(a\approx0.9994\) c) \(T_{1/2}\approx1205\,\text{years}\) d) About \(8004\) years
Half-life and doubling time models 88
Air pressure decreases approximately exponentially as altitude increases. For each \(1000\,\text{m}\) increase in altitude, air pressure decreases by about \(12\%\). a) What percent of sea-level air pressure remains at an altitude of \(3000\,\text{m}\)? Round to the nearest tenth of a percent. b) Find the pressure factor for altitude increases of \(1000\,\text{m}\), \(500\,\text{m}\), and \(5000\,\text{m}\). Round nonexact factors to four decimal places. c) At approximately what altitude is the air pressure one-half of its sea-level value? Round to the nearest meter.

Hints

- Write a pressure model using altitude measured in thousands of meters. - A \(500\,\text{m}\) increase is one-half of the standard \(1000\,\text{m}\) step. - For the half-pressure altitude, set the remaining fraction equal to \(0.5\).

Solution

1. The factor per \(1000\,\text{m}\) is \(1-0.12=0.88\). 2. At \(3000\,\text{m}\), the remaining fraction is \(0.88^3=0.681472\), or approximately \(68.1\%\). 3. The factors are \(0.88\) for \(1000\,\text{m}\), \(\sqrt{0.88}\approx0.9381\) for \(500\,\text{m}\), and \(0.88^5\approx0.5277\) for \(5000\,\text{m}\). 4. Let \(x\) be altitude measured in thousands of meters. From \(0.88^x=0.5\), obtain \(x=\frac{\ln(0.5)}{\ln(0.88)}\approx5.42227\). Thus, altitude is approximately \(5422\,\text{m}\).

Answer

a) About \(68.1\%\) b) \(1000\,\text{m}\): \(0.88\); \(500\,\text{m}\): \(0.9381\); \(5000\,\text{m}\): \(0.5277\) c) About \(5422\,\text{m}\)
Half-life and doubling time models 89
The Rule of 72 estimates the doubling time \(T_d\), in years, of an investment growing at an annual rate of \(p\%\): \(T_d\approx\frac{72}{p}\). Use the exponential growth model between whole years as well. 1. An initial investment of \(\$5000\) grows at \(3\%\) per year. Find its doubling time to the nearest hundredth of a year and compare it with the Rule of 72 estimate. 2. For \(p=12\), find the percent error of the Rule of 72 estimate relative to the logarithmic doubling time. Round to the nearest hundredth of a percent. 3. Show that \(T_d=\frac{\ln(2)}{\ln\left(1+\frac{p}{100}\right)}\). Calculate \(pT_d\) for annual rates of \(1\%\) and \(5\%\), to the nearest hundredth. Explain why a number near \(70\) is useful for estimating doubling time at small growth rates.

Hints

- What equation describes an investment doubling from its initial balance? - Which value belongs in the denominator of a percent-error calculation? - Take logarithms to solve the doubling equation for time. - For the final comparison, multiply each computed doubling time by its numerical percentage rate.

Solution

1. The doubling equation is \(5000(1.03)^{T_d}=10{,}000\), giving \(T_d=\frac{\ln(2)}{\ln(1.03)}\approx23.45\) years. The Rule of 72 gives \(24\) years, about \(0.55\) year longer. 2. For \(p=12\), the logarithmic doubling time is \(T_d=\frac{\ln(2)}{\ln(1.12)}\approx6.1163\) years; the estimate is \(6\) years. The relative error is \(\frac{|6-6.1163\dots|}{6.1163\dots}\cdot100\%\approx1.90\%\). 3. From \(\left(1+\frac{p}{100}\right)^{T_d}=2\), take logarithms to obtain \(T_d\ln\left(1+\frac{p}{100}\right)=\ln(2)\), so \(T_d=\frac{\ln(2)}{\ln\left(1+\frac{p}{100}\right)}\). 4. For \(p=1\), \(pT_d=\frac{\ln(2)}{\ln(1.01)}\approx69.66\). For \(p=5\), \(pT_d=\frac{5\ln(2)}{\ln(1.05)}\approx71.03\). Both products are near \(70\), which explains why dividing a fixed number around \(70\) by a small percentage rate gives a useful estimate.

Answer

1. \(T_d\approx23.45\) years; Rule of 72 estimate: \(24\) years 2. About \(1.90\%\) 3. \(T_d=\frac{\ln(2)}{\ln\left(1+\frac{p}{100}\right)}\); \(pT_d\approx69.66\) for \(1\%\), and \(pT_d\approx71.03\) for \(5\%\), so a constant near \(70\) gives a useful estimate for small rates.

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