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Half-life and doubling time models

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Difficulty: 1 – 5

55149212

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

\(40\,\text{mg}\)

55149312

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

52645512

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

52833512

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

52833612

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

52841712

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

52848812

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

52849212

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

53449212

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) \(A(t)=5\cdot2^{t/4}\); \(A(10)\approx28.28\,\text{m}^2\)

51008912

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) \(33.2\,\text{min}\)

52641612

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. \(T_{1/2}\approx5.9\,\text{hours}\)

52645612

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

52646112

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) \(T_{1/2}\approx34.31\)

52646212

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) \(t\approx83.05\,\text{hours}\)

52646412

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.

52647312

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 \(19.81\,\text{hours}\)

52647412

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 \(39.24\,\text{years}\)

52649812

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 \(26.87\,\text{years}\)

52670212

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) \(T_d\approx4.80\,\text{hours}\)

52824512

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

52824612

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) \(b\approx0.871\)

52831312

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) \(2\%\)

52831412

Under ideal conditions, a bacteria 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

- Convert days to hours. - A doubling time of \(14\) hours means the \(14\)-hour growth factor is \(2\). - Rearrange the Rule of 70 to estimate the rate. - Round the population 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

52832512

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

52832612

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

52832912

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

52833012

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: \(T_d\approx25.1\,\text{years}\)

52833212

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 \(6.4\%\)

52833412

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

52833712

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

52833812

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

52834312

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 \(9.92\%\)

52834412

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

52834612

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

52836112

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 \(0.64\,\text{mg}\)

52848712

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) \(10\,\text{mg}\)

52849012

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

52852912

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 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) The medication is considered mostly eliminated when less than \(50\,\text{mg}\) remains. After how many complete hours does this first occur?

Hints

- Convert the hourly percent decrease to a decay factor. - Use the same factor repeatedly to fill the table. - At the half-life, one half of the initial amount remains. - Use logarithms to find the boundary time, then check whole hours.

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. Find the boundary by solving \(400(0.85)^t=50\), or \(0.85^t=0.125\). Then \(t\approx12.80\) hours. 5. After \(12\) hours, about \(56.90\,\text{mg}\) remains; after \(13\) hours, about \(48.36\,\text{mg}\) remains. Therefore, the amount first falls below \(50\,\text{mg}\) after \(13\) complete 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

52853012

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

52853112

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 \(190.22\,\text{Bq}\)

52853212

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

52853712

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) \(T_{1/2}\approx28.5\,\text{years}\)

52853812

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

52982612

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) The procedure may end when the activity has fallen to \(5\%\) of its initial value. Find how many hours after injection this occurs. Round to the nearest hundredth of an hour.

Hints

- Use the activity after \(4\) hours to determine the continuous decay constant. - At the half-life, the exponential factor is one half. - For part c), compare the remaining activity with the initial activity as a ratio. - Use logarithms when the unknown is in an exponent.

Solution

1. Using \(A(4)=240\), \(240=400e^{4k}\), so \(e^{4k}=0.6\). 2. Therefore, \(k=\frac{\ln(0.6)}{4}\approx-0.1277\). 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) \(k\approx-0.1277\) b) \(T_{1/2}\approx5.43\,\text{hours}\) c) About \(23.46\,\text{hours}\)

52982812

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 \(9.97\,\text{hours}\)

52983412

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.

52983912

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 population change in exactly one hour? Give the exact factor and its value to four decimal places. c) By what factor does it change in \(4.5\) 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

- Translate the doubling condition into an equation involving the hourly factor. - A change factor over several hours is a power of the one-hour factor. - Convert \(30\) minutes to hours before using the model. - To convert a growth factor to a percent increase, compare it with \(1\).

Solution

1. The doubling condition gives \(b^3=2\), so \(b=2^{1/3}\approx1.2599\). 2. The one-hour factor is \(b=2^{1/3}\approx1.2599\). 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) \(2^{1/3}\approx1.2599\) c) \(\sqrt{8}\approx2.8284\) d) About \(12.25\%\)

52984612

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) \(T_{1/2}\approx3.11\,\text{hours}\)

52993912

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 \(8.5\%\)

52994012

A threatened animal population in a reserve is modeled by \(B(t)=B_0e^{a(t-t_0)}\). In 2015, the reserve had \(1200\) animals, and the continuous growth rate is \(a=0.045\,\text{yr}^{-1}\). a) Find the model's predicted population in 2030. Round to the nearest whole animal. b) Determine during which year the population reaches \(2400\). 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

- Find the elapsed time from 2015 to the target year. - A doubled population makes the exponential growth factor equal to \(2\). - The effective one-year growth factor is obtained by evaluating the exponential factor over one year. - Compare that one-year factor with \(1\) to obtain a percent increase.

Solution

1. From 2015 to 2030 is \(15\) years. Thus, \(B(2030)=1200e^{0.045(15)}\approx2356.84\), which rounds to \(2357\) animals. 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 \(e^{0.045}\approx1.0460\), giving an effective annual growth rate of \((e^{0.045}-1)\cdot100\%\approx4.60\%\). 4. This exceeds \(4.5\%\) because \(e^x>1+x\) for \(x>0\).

Answer

a) About \(2357\) animals b) During 2030 c) About \(4.60\%\) per year; the effective annual rate exceeds the continuous rate's numerical percentage.

52994412

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 \(27\,\text{years}\)

53002512

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 \(96.05\,\text{days}\)

53002612

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.

53002912

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 \(9.60\,\text{years}\)

53003212

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.

53003412

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

53003512

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 has an initial dose of \(300\,\text{mg}\) and decay constant \(k=-0.12\,\text{h}^{-1}\). Write its model in the form \(g(t)=be^{kt}\), and compare its graph with the graph of \(M\).

Hints

- Match the discrete decay factor with an equivalent base-\(e\) factor. - At the half-life, one half of the initial amount remains. - For the graph comparison, consider both the initial values and the magnitudes of the negative decay constants.

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. Because its decay constant has smaller magnitude, \(g\) 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 remains above the graph of \(M\).

Answer

a) \(k\approx-0.1985\,\text{h}^{-1}\) b) \(T_{1/2}\approx3.49\,\text{hours}\) c) \(g(t)=300e^{-0.12t}\); it starts higher, remains above \(M\), and decreases more slowly proportionally.

53218412

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) \(2.5\,\text{mg}\)

53245712

The graph shows the concentration \(C(t)\) of a medication in Elena's blood, in milligrams per liter, \(\text{mg/L}\). Time \(t\) is measured in hours after the dose. 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) Elena may take another dose only after the concentration falls below \(0.1\,\text{mg/L}\). Find when this occurs. Round to the nearest tenth of an hour.

Figure for problem 532457

Hints

- Read the initial concentration from the marked point at \(t=0\). - Compare the marked concentrations at equal time intervals. - Use one nonzero-time point to determine the hourly decay factor. - For the threshold, compare the target concentration with the initial concentration before taking logarithms.

Solution

1. The point \((0, 8)\) gives \(C_0=8\). 2. The marked values are halved 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 \(a=(0.5)^{1/3}\approx0.7937\). Thus, \(C(t)=8(0.7937)^t\). 4. Solve \(8(0.7937)^t=0.1\). Then \(t=\frac{\ln(0.0125)}{\ln(0.7937)}\approx19.0\) hours. The concentration is below the limit just after this time.

Answer

a) \(C(t)=8(0.7937)^t\); \(T_{1/2}=3\,\text{hours}\) b) Just after about \(19.0\,\text{hours}\)

53262612

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) \(f(15)=320\); \(g(15)\approx5.9\)

53402812

Iodine-131 is used in some thyroid treatments. The graph shows the decay of a sample with an initial mass of \(10\,\text{mg}\). a) Determine the half-life \(T_{1/2}\) from the graph. 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

- Find the mass that is one half of the initial mass. - Read from the graph when the curve reaches that mass. - Use the characteristic-time form of an exponential decay model. - Substitute the requested time into the model for the last part.

Solution

1. Half the initial mass is \(5\,\text{mg}\). The graph reaches \(5\,\text{mg}\) at \(t=8\) days, so \(T_{1/2}=8\) days. 2. Substituting \(m_0=10\) and \(T_{1/2}=8\) gives \(m(t)=10\cdot0.5^{t/8}\). 3. After \(20\) days, \(m(20)=10\cdot0.5^{20/8}\approx1.77\,\text{mg}\).

Answer

a) \(T_{1/2}=8\,\text{days}\) b) \(m(t)=10\cdot0.5^{t/8}\) c) Approximately \(1.77\,\text{mg}\)

52594812

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 an exponential growth model for each culture. - Express the number of doubling or tripling periods as a fraction of the unknown time. - Set each model equal to the target population and solve for time. - Use logarithms to isolate a variable in an exponent.

Solution

1. Let \(t\) be time in minutes. The models are \(N_A(t)=100\cdot2^{\frac{t}{20}}\) and \(N_B(t)=100\cdot3^{\frac{t}{30}}\). 2. For Culture A, solve \(100\cdot2^{\frac{t}{20}}=1{,}000{,}000\). Then \(2^{\frac{t}{20}}=10{,}000\), so \(t_A=20\cdot\frac{\ln(10{,}000)}{\ln(2)}\approx265.75\) minutes. 3. For Culture B, solve \(100\cdot3^{\frac{t}{30}}=1{,}000{,}000\). Then \(3^{\frac{t}{30}}=10{,}000\), so \(t_B=30\cdot\frac{\ln(10{,}000)}{\ln(3)}\approx251.51\) minutes. 4. Culture B reaches the target first. The time difference is \(265.75-251.51\approx14.25\) minutes.

Answer

Culture B reaches \(1{,}000{,}000\) bacteria first, about \(14.25\) minutes before Culture A.

52834512

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

52835812

In a lake, light intensity decreases exponentially with water depth. The intensity decreases by \(22\%\) for each meter of depth. Let the intensity at the surface be \(100\%\). a) What percent of the surface light intensity remains at depths of \(2\,\text{m}\), \(5\,\text{m}\), and \(10\,\text{m}\)? b) At what depth is the light intensity one-half of the surface intensity? c) The aphotic zone begins where the light intensity falls below \(1\%\) of the surface intensity. Find the depth at which the intensity is \(1\%\).

Hints

- Convert the percent decrease to a decay factor. - Use an exponential model with depth as the input. - To find a depth for a specified intensity, solve an exponential equation. - You may model the surface intensity as either \(100\) or \(1\), provided you use the corresponding target value.

Solution

1. The decay factor per meter is \(1-0.22=0.78\), so the model is \(I(d)=100(0.78)^d\), where \(d\) is depth in meters and \(I(d)\) is a percent. 2. \(I(2)=100\cdot(0.78)^2=60.84\%\), \(I(5)=100\cdot(0.78)^5\approx28.87\%\), and \(I(10)=100\cdot(0.78)^{10}\approx8.34\%\). 3. For half the surface intensity, solve \(0.78^d=0.5\): \(d=\frac{\ln(0.5)}{\ln(0.78)}\approx2.79\,\text{m}\). 4. For \(1\%\) intensity, solve \(0.78^d=0.01\): \(d=\frac{\ln(0.01)}{\ln(0.78)}\approx18.53\,\text{m}\).

Answer

a) At \(2\,\text{m}\): \(60.84\%\); at \(5\,\text{m}\): about \(28.87\%\); at \(10\,\text{m}\): about \(8.34\%\) b) About \(2.79\,\text{m}\) c) About \(18.53\,\text{m}\)

53003312

The Rule of 72 is a common estimate for the doubling time \(T_d\), in years, of an investment earning an annual interest rate of \(p\%\): \(T_d\approx\frac{72}{p}\). 1. An initial investment of \(\$5000\) earns \(3\%\) interest compounded annually. Find the exact-model 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 exact-model doubling time. Round to the nearest hundredth of a percent. 3. Show that the exact doubling time is \(T_d=\frac{\ln 2}{\ln\left(1+\frac{p}{100}\right)}\). Then use \(\ln(1+x)\approx x\) for small \(x\) to explain why \(T_dp\approx69.3\) provides a theoretical basis for doubling-time rules.

Hints

- Set the final balance equal to twice the initial balance. - Compare the Rule of 72 estimate with the logarithmic solution of the compound-interest equation. - Percent error compares the absolute difference with the exact-model value. - In the derivation, take logarithms before applying the supplied small-\(x\) approximation.

Solution

1. The doubling equation is \(5000(1.03)^{T_d}=10{,}000\), so \((1.03)^{T_d}=2\). Therefore, \(T_d=\frac{\ln2}{\ln(1.03)}\approx23.45\) years. The Rule of 72 gives \(\frac{72}{3}=24\) years, about \(0.55\) year longer. 2. For \(p=12\), the exact-model doubling time is \(T_d=\frac{\ln2}{\ln(1.12)}\approx6.1163\) years, while the estimate is \(6\) years. The percent error is \(\frac{|6-6.1163|}{6.1163}\cdot100\%\approx1.90\%\). 3. From \(\left(1+\frac{p}{100}\right)^{T_d}=2\), taking natural logarithms gives \(T_d\ln\left(1+\frac{p}{100}\right)=\ln2\). Hence \(T_d=\frac{\ln2}{\ln\left(1+\frac{p}{100}\right)}\). 4. For small \(p\), \(\ln\left(1+\frac{p}{100}\right)\approx\frac{p}{100}\), so \(T_d\approx\frac{100\ln2}{p}\). Since \(100\ln2\approx69.3\), \(T_dp\approx69.3\).

Answer

1. \(T_d\approx23.45\,\text{years}\); Rule of 72 estimate: \(24\,\text{years}\) 2. About \(1.90\%\) 3. \(T_d=\frac{\ln2}{\ln\left(1+\frac{p}{100}\right)}\), and for small \(p\), \(T_dp\approx69.3\)

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