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Exponential growth and decay functions

Learning goals

  • Recognizes exponential growth or decay from repeated multiplicative change.
  • Writes an exponential model with an initial value and growth or decay factor.
  • Interprets parameters in context.
  • Evaluates the model at specified inputs.
  • Distinguishes percent change per interval from absolute change.

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Exponential growth and decay functions 1
Consider the exponential function \(f(x)=6(1.3)^x\). State the initial value, the factor for an increase of \(1\) in \(x\), and whether the function shows growth or decay.

Hints

- What does an exponent of \(0\) do to the exponential factor? - In the form \(a b^x\), which number controls the change when \(x\) increases by \(1\)? - Compare the factor with \(1\).

Solution

1. At \(x=0\), \(f(0)=6(1.3)^0=6\), so the initial value is \(6\). 2. Each increase of \(1\) in \(x\) multiplies the output by \(1.3\). 3. Because \(1.3>1\), the function shows exponential growth.

Answer

Initial value: \(6\) Factor: \(1.3\) Exponential growth
Exponential growth and decay functions 2
An exponential quantity starts at \(40\) and doubles each hour. Write a function \(Q(t)\) for the quantity after \(t\) hours.

Hints

- Identify the value at \(t=0\). - Translate “doubles” into a multiplication factor.

Solution

1. The initial value is \(40\). 2. Doubling means a growth factor of \(2\) for each hour. 3. Therefore, \(Q(t)=40(2)^t\).

Answer

\(Q(t)=40(2)^t\)
Exponential growth and decay functions 3
The table represents an exponential function. <table><tr><th>\(x\)</th><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr><tr><th>\(f(x)\)</th><td>\(7\)</td><td>\(14\)</td><td>\(28\)</td><td>\(\square\)</td></tr></table> Find the missing output.

Hints

- Compare two consecutive known outputs. - Apply the same multiplicative change one more time.

Solution

1. Each output is twice the preceding output. 2. Therefore, the missing output is \(28\cdot2=56\).

Answer

\(56\)
Exponential growth and decay functions 4
Let \(f(x)=20(0.75)^x\). Find \(f(1)\).

Hints

- Substitute the given input into the function. - An exponent of \(1\) leaves the factor unchanged.

Solution

1. Substitute \(x=1\): \(f(1)=20(0.75)^1\). 2. Thus, \(f(1)=15\).

Answer

\(15\)
Exponential growth and decay functions 5
Two number patterns start at \(5\). - Pattern A: \(5,10,20,40\) - Pattern B: \(5,10,15,20\) Which pattern shows exponential growth? Give its step-by-step growth factor and explain your choice.

Hints

- Compare how consecutive numbers change in each pattern. - Does one pattern repeatedly add the same amount while the other repeatedly multiplies by the same factor?

Solution

1. Pattern A has the constant ratio \(\frac{10}{5}=\frac{20}{10}=\frac{40}{20}=2\). 2. Pattern B adds \(5\) each step; its ratios are not constant. 3. Pattern A is exponential, with a growth factor of \(2\) per step.

Answer

Pattern A; growth factor \(2\). Each term is twice the previous one, while Pattern B adds \(5\).
Exponential growth and decay functions 6
This table follows an exponential pattern. Fill in the missing output and explain how you found it. <table> <tr><th>\(x\)</th><th>\(-1\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>\(f(x)\)</th><td>\(48\)</td><td>\(24\)</td><td>\(12\)</td><td>\(\square\)</td></tr> </table>

Hints

- Compare neighboring entries, not just the first and last. - Does the same multiplication work for both known steps?

Solution

1. Moving one column to the right halves the output: \(\frac{24}{48}=\frac{12}{24}=\frac{1}{2}\). 2. Halving again gives \(f(2)=6\). 3. The fixed factor \(\frac{1}{2}\) makes this exponential decay, not a fixed subtraction.

Answer

\(f(2)=6\), because each output is multiplied by \(\frac{1}{2}\).
Exponential growth and decay functions 7
a) What is \(f(0)\)? b) By what factor does the output change when \(x\) increases by \(1\)?
Figure for problem 560282

Hints

- Find the point where \(x=0\). - Compare the heights at neighboring whole-number inputs. - Is the output being increased by a fixed amount or multiplied by a fixed factor?

Solution

1. The curve meets the y-axis at \(2\), so \(f(0)=2\). 2. The marked outputs at \(x=1\) and \(x=2\) are \(4\) and \(8\). 3. Each one-unit increase multiplies the output by \(2\).

Answer

a) \(f(0)=2\) b) Factor \(2\)
Exponential growth and decay functions 8
Which curve shows exponential growth and which shows exponential decay? Explain how you know.
Figure for problem 560283

Hints

- Follow each curve from left to right. - What does an increasing output tell you about growth or decay?

Solution

1. Curve \(p\) rises as \(x\) increases, so it represents exponential growth. 2. Curve \(q\) falls while staying above the x-axis, so it represents exponential decay.

Answer

\(p\) shows growth; \(q\) shows decay. Curve \(p\) rises to the right, while \(q\) falls to the right.
Exponential growth and decay functions 9
Rowan sees the pattern \(40,60,90,135\) and says, “It goes up by \(20\) each step.” Explain the mistake and give the correct rule for moving to the next term.

Hints

- Check the change from the second term to the third. - Try dividing each term by the one before it.

Solution

1. The increases are \(20\), \(30\), and \(45\), so adding \(20\) does not work after the first step. 2. The consecutive ratios are all \(1.5\), so each term is the previous term multiplied by \(1.5\).

Answer

The increases are not constant. Multiply by \(1.5\) each step, rather than adding \(20\).
Exponential growth and decay functions 10
A bacterial culture starts with \(1200\) bacteria. After \(5\) hours, it has \(3000\) bacteria. Assume the population grows exponentially. Write a function \(N(t)\) for the population after \(t\) hours. Use an exact expression; do not round the growth factor.

Hints

- What is the starting population? - Find the factor for the whole five-hour interval first. - How many five-hour intervals fit into \(t\) hours?

Solution

1. The initial population is \(1200\). 2. In \(5\) hours, the population is multiplied by \(\frac{3000}{1200}=2.5\). 3. Since each five-hour interval has the same factor, the number of five-hour intervals in \(t\) hours is \(\frac{t}{5}\). Thus \(N(t)=1200(2.5)^{t/5}\).

Answer

\(N(t)=1200(2.5)^{t/5}\)
Exponential growth and decay functions 11
Consider three population models, with \(t\) measured in years: (1) \(f_1(t)=600(1.05)^t\) (2) \(f_2(t)=450(0.88)^t\) (3) \(f_3(t)=600(0.95)^t\) a) Which population is smallest at \(t=0\)? Give its size. b) Which models describe decay? Explain how you know. c) Find \(f_1(20)\). Round to the nearest hundredth.

Hints

- What does an exponent of \(0\) do to each base? - Compare the three bases with \(1\). - Use the given first model to find its value at year \(20\).

Solution

1. At \(t=0\), each exponential factor equals \(1\), so the starting sizes are \(600\), \(450\), and \(600\). Model (2) starts at \(450\). 2. A positive base between \(0\) and \(1\) makes an exponential model decrease. Models (2) and (3) have bases \(0.88\) and \(0.95\). 3. \(f_1(20)=600(1.05)^{20}\approx1591.98\).

Answer

a) Model (2): \(450\) b) Models (2) and (3), since both bases are between \(0\) and \(1\). c) \(f_1(20)\approx1591.98\)
Exponential growth and decay functions 12
A quantity follows a model of the form \(B(x)=ca^x\). a) Identify \(c\) and \(a\) for \(B(x)=15.5(0.6)^x\). b) Does the quantity grow or decay as \(x\) increases? Explain. c) Find \(B(5)\). Round to two decimal places.

Hints

- Match the given model to \(ca^x\). - Think about repeated multiplication by a factor less than \(1\). - Substitute \(5\) for \(x\), then round at the end.

Solution

1. Comparing the model with \(ca^x\) gives \(c=15.5\) and \(a=0.6\). 2. Since \(0<0.6<1\), the quantity is multiplied by a factor less than \(1\) at each step, so it decays. 3. \(B(5)=15.5(0.6)^5=1.20528\approx1.21\).

Answer

a) \(c=15.5\), \(a=0.6\) b) Decay, since \(0<a<1\). c) \(B(5)\approx1.21\)
Exponential growth and decay functions 13
A wildlife preserve has \(40\) deer at the beginning of a study. After one year, the population has grown to \(48\) deer. Assume the population grows exponentially each year. Write a function \(f(t)\) that models the deer population after \(t\) years.

Hints

- Identify the population at \(t=0\). - Compare the population at year \(1\) with the initial population. - Use the general form of an exponential function. - Divide the year-1 value by the initial value to find the growth factor.

Solution

1. The initial value is \(40\), so \(f(t)=40b^t\). 2. Use the value after one year: \(48=40b\), so \(b=\frac{48}{40}=1.2\). 3. Therefore, \(f(t)=40(1.2)^t\).

Answer

\(f(t)=40(1.2)^t\)
Exponential growth and decay functions 14
In an exponential process, the amount \(f(t)\) is multiplied by \(4\) whenever time increases by \(2\) units. Write an equation expressing \(f(t-2)\) using only \(f(t)\) and a numerical factor.

Hints

- What relationship links \(f(t)\) and \(f(t+2)\)? - How can you undo a multiplication? - You do not need to find the change for one time unit.

Solution

1. The given relationship is \(f(t+2)=4f(t)\). 2. To express the amount two units earlier, reverse this change: \(f(t-2)=\frac{1}{4}f(t)\).

Answer

\(f(t-2)=\frac{1}{4}f(t)\)
Exponential growth and decay functions 15
The number of users of a new app doubles each month. At the beginning of the study, the app has \(4500\) users. Write an exponential function \(N(t)\) for the number of users after \(t\) months. Then find the number of users after half a year and after one year.

Hints

- Identify the number of users at \(t=0\). - Translate “doubles” into a growth factor. - Convert half a year and one year to months before substituting for \(t\).

Solution

1. The initial value is \(N(0)=4500\), and monthly doubling gives a growth factor of \(2\). Therefore, \(N(t)=4500(2)^t\). 2. Half a year is \(6\) months: \(N(6)=4500\cdot2^6=4500\cdot64=288{,}000\). 3. One year is \(12\) months: \(N(12)=4500\cdot2^{12}=4500\cdot4096=18{,}432{,}000\).

Answer

The function is \(N(t)=4500(2)^t\). After half a year: \(288{,}000\) users After one year: \(18{,}432{,}000\) users
Exponential growth and decay functions 16
A protected plant population grows by a factor of \(1.4\) each year. At the beginning of the study, there are \(250\) plants. Write a function \(B(t)\) for the population after \(t\) years. Then find the predicted population after \(5\) years and after \(10\) years. Round to the nearest whole plant.

Hints

- Identify the initial value and the yearly multiplication factor. - Substitute each requested time into the model. - Round population predictions to whole plants.

Solution

1. The initial value is \(250\), and the yearly growth factor is \(1.4\), so \(B(t)=250(1.4)^t\). 2. After \(5\) years, \(B(5)=250\cdot(1.4)^5=1344.56\), or about \(1345\) plants. 3. After \(10\) years, \(B(10)=250\cdot(1.4)^{10}\approx7231.37\), or about \(7231\) plants.

Answer

\(B(t)=250(1.4)^t\) After \(5\) years: about \(1345\) plants After \(10\) years: about \(7231\) plants
Exponential growth and decay functions 17
The approximation \(2^{10}\approx 10^3\) is often used in computing and engineering to estimate powers of \(2\). a) Find the exact value of \(2^{10}\). By what percent does the exact value exceed \(1000\), using \(1000\) as the reference value? b) Suppose computing power doubles every \(2\) years. Use the approximation to explain why computing power increases by about a factor of \(10^3\) in \(20\) years and by about a factor of \(10^9\) in \(60\) years. c) A computer currently performs \(10^{16}\) operations per second. Estimate its performance after \(40\) years under the same growth assumption. Write the result in scientific notation.

Hints

- Divide the elapsed time by the doubling time to find the number of doublings. - Rewrite large powers using multiples of \(10\) in the exponent. - For part a, compare the excess \(1024-1000\) with the stated reference value \(1000\).

Solution

1. \(2^{10}=1024\). Relative to \(1000\), the exact value exceeds the reference by \(\frac{1024-1000}{1000}(100\%)=2.4\%\). 2. Twenty years contains \(10\) doubling periods, so the factor is \(2^{10}\approx10^3\). Sixty years contains \(30\) doubling periods, so \(2^{30}=(2^{10})^3\approx(10^3)^3=10^9\). 3. Forty years contains \(20\) doubling periods. Thus, the factor is \(2^{20}=(2^{10})^2\approx10^6\), and the estimated performance is \(10^{16}\cdot10^6=10^{22}=1\times10^{22}\) operations per second.

Answer

a) \(2^{10}=1024\); \(1024\) exceeds \(1000\) by \(2.4\%\) b) \(2^{10}\approx10^3\) in \(20\) years; \(2^{30}\approx10^9\) in \(60\) years c) \(1\times10^{22}\) operations per second
Exponential growth and decay functions 18
Find the base \(a\) of the exponential function \(f(x)=a^x\) if its graph passes through \((-2, 6.25)\).

Hints

- Substitute the point into the function. - Rewrite the negative exponent as a reciprocal. - Remember that an exponential base must be positive.

Solution

1. Substitute the point: \(6.25=a^{-2}\). 2. Since \(6.25=\frac{25}{4}\), \(a^2=\frac{4}{25}\). 3. An exponential base must be positive, so \(a=\frac{2}{5}=0.4\).

Answer

\(a=0.4\)
Exponential growth and decay functions 19
Exponential functions of the form \(f(x)=b^x\) require restrictions on the base. 1. Explain the problem with evaluating \(f(0.5)\) when \(b=-4\). What general restriction on \(b\) ensures that \(f\) is defined for every real input? 2. Why is \(b=1\) usually excluded from the definition of an exponential function? Describe the function that results in this case. 3. Show algebraically that every graph in the family \(f(x)=b^x\), for every allowed base \(b\), passes through the same point on the y-axis.

Hints

- Rewrite an exponent of \(0.5\) as a square root. - Substitute \(b=1\) and compare several function values. - How do you find a y-intercept? - Recall the zero-exponent rule.

Solution

1. \(f(0.5)=(-4)^{0.5}=\sqrt{-4}\), which is not a real number. To define \(b^x\) for every real \(x\), the base must satisfy \(b>0\). 2. If \(b=1\), then \(f(x)=1^x=1\) for every \(x\). This is a constant function and does not represent exponential growth or decay. 3. A y-intercept occurs at \(x=0\). For every allowed base, \(f(0)=b^0=1\). Therefore, every graph passes through \((0, 1)\).

Answer

1. \((-4)^{0.5}=\sqrt{-4}\) is not real, so \(b>0\) is required. 2. When \(b=1\), \(f(x)=1\), a constant function. 3. Since \(b^0=1\), all graphs pass through \((0, 1)\).
Exponential growth and decay functions 20
The graph of \(f(x)=ab^x\), where \(a,b>0\), passes through \((2, 5)\) and \((3, 20)\). a) Find \(a\) and \(b\). b) Another function \(g(x)=cd^x\) passes through \((2, 5)\) and \((3, 15)\). Without fully calculating \(d\), explain whether \(d\) is greater than or less than \(b\).

Hints

- Divide function values whose inputs differ by \(1\). - Use one point after finding the base. - Compare the one-step growth factors of the two functions.

Solution

1. The ratio of consecutive function values is the base: \(b=\frac{f(3)}{f(2)}=\frac{20}{5}=4\). 2. Use \(f(2)=5\): \(5=a(4)^2\), so \(a=\frac{5}{16}=0.3125\). 3. Both functions equal \(5\) at \(x=2\), but one unit later \(g(3)=15<20=f(3)\). Therefore, \(d=\frac{15}{5}<\frac{20}{5}=b\), so \(d<b\).

Answer

a) \(a=0.3125\), \(b=4\) b) \(d<b\)
Exponential growth and decay functions 21
A bacterial culture grows under ideal conditions. a) Estimate the doubling time to the nearest half-hour. b) About when does the population reach \(400\) bacteria? Give your answer to the nearest half-hour. c) Estimate the population after \(9\) hours, to the nearest \(50\) bacteria.
Figure for problem 533190

Hints

- Find the population at \(t=0\) on the graph before looking for the doubling time. - To find a time, start from the population on the vertical scale. - To find a population, start from the time on the horizontal scale.

Solution

1. At \(t=0\), the graph starts at about \(100\) bacteria. 2. The curve reaches \(200\) bacteria at about \(3.1\) hours, or \(3\) hours to the nearest half-hour. 3. The curve reaches \(400\) bacteria at about \(6.2\) hours, or \(6\) hours to the nearest half-hour. 4. At \(t=9\), the curve is close to \(750\), so the population is about \(750\) bacteria.

Answer

a) About \(3\) hours b) About \(6\) hours c) About \(750\) bacteria
Exponential growth and decay functions 22
A cup of hot tea cools in a room at \(70\,\text{°F}\). In this model, the difference between the tea's temperature and the room temperature decreases by the same percent each minute. a) What was the tea's temperature at the beginning? b) About how long does it take to cool to \(105\,\text{°F}\)? c) Why does the tea get closer to room temperature without cooling below it in this model?
Figure for problem 533253

Hints

- Read the value where the graph meets the vertical axis. - Find the height \(105\) on the temperature scale and read its time. - What would a temperature difference of \(0\) mean here?

Solution

1. The graph starts at \(180\,\text{°F}\). 2. The curve reaches about \(105\,\text{°F}\) at \(t=10\) minutes. 3. The initial difference from the room is positive. Each minute it is multiplied by the same factor between \(0\) and \(1\), so it shrinks but remains positive. The tea stays above \(70\,\text{°F}\) in this model.

Answer

a) \(180\,\text{°F}\) b) About \(10\) minutes c) The temperature difference starts positive. Each minute it is multiplied by a factor between \(0\) and \(1\), so it gets smaller but stays positive. The tea stays above \(70\,\text{°F}\).
Exponential growth and decay functions 23
Match equations (a) through (d) with graphs \(p\), \(q\), \(r\), and \(s\). Justify each match using the y-intercept and growth or decay behavior. (a) \(y=3(1.4)^x\) (b) \(y=3(0.6)^x\) (c) \(y=1.5(1.4)^x\) (d) \(y=1.5(0.6)^x\)
Figure for problem 533504

Hints

- Evaluate each function at \(x=0\) to find its y-intercept. - Compare each base with \(1\). - For graphs with the same initial value, distinguish growth from decay.

Solution

1. Equations (a) and (b) have y-intercept \(3\), matching graphs \(p\) and \(q\). Equations (c) and (d) have y-intercept \(1.5\), matching graphs \(r\) and \(s\). 2. A base greater than \(1\) produces growth, while a base between \(0\) and \(1\) produces decay. 3. Therefore, (a) matches \(p\), (b) matches \(q\), (c) matches \(r\), and (d) matches \(s\).

Answer

(a) – \(p\); (b) – \(q\); (c) – \(r\); (d) – \(s\)
Exponential growth and decay functions 24
The graph shows two bacterial cultures modeled by \(N(t)=N_0a^t\), where \(t\) is measured in hours and \(N(t)\) is measured in thousands of bacteria. Find \(N_0\) and \(a\) for each culture.
Figure for problem 534388

Hints

- Read the initial value at \(t=0\). - The one-hour growth factor is the ratio \(N(1)/N(0)\). - Use clearly marked grid points.

Solution

1. At \(t=0\), the graph gives \(N_0=2\) for culture \(h\) and \(N_0=8\) for culture \(k\). 2. At \(t=1\), the graph gives \(h(1)=3\) and \(k(1)=6\). 3. For culture \(h\), \(2a=3\), so \(a=1.5\). For culture \(k\), \(8a=6\), so \(a=0.75\).

Answer

Culture \(h\): \(N_0=2\) thousand, \(a=1.5\) Culture \(k\): \(N_0=8\) thousand, \(a=0.75\)
Exponential growth and decay functions 25
The table represents an exponential function. Find the missing output and state the factor for each increase of \(1\) in \(x\). <table> <tr><th>\(x\)</th><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><th>\(f(x)\)</th><td>\(12\)</td><td></td><td>\(27\)</td><td>\(40.5\)</td></tr> </table>

Hints

- Use two consecutive known outputs to determine the multiplicative change. - Once the factor is known, work one step backward to the missing value. - Check that the same factor carries the missing value forward to the next known output.

Solution

1. From \(x=2\) to \(x=3\), the factor is \(\frac{40.5}{27}=1.5\). 2. The same factor applies at every one-unit step, so \(f(1)=\frac{27}{1.5}=18\).

Answer

Missing output: \(18\) Factor: \(1.5\)
Exponential growth and decay functions 26
Eli claims that \(f(x)=10(0.6)^x\) eventually becomes negative because the number keeps shrinking. a) Find \(f(2)\). b) Explain why Eli's claim is false for every real \(x\).

Hints

- Try evaluating the rule at a few larger whole-number inputs. What happens to the sign? - What do positive bases do when raised to negative or fractional exponents? - Think about whether getting closer to zero is the same as reaching or passing it.

Solution

1. \(f(2)=10(0.6)^2=3.6\). 2. The positive base \(0.6\) raised to any real exponent is positive. Multiplying by \(10\) keeps the output positive. 3. The outputs can become small but never negative.

Answer

a) \(3.6\) b) The output is always positive because \(10>0\) and \((0.6)^x>0\) for every real \(x\).
Exponential growth and decay functions 27
A runner records the distance she runs each week. The table gives her weekly totals. <table> <tr><th>Week \(t\)</th><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td></tr> <tr><th>Miles \(D(t)\)</th><td>\(8\)</td><td>\(10\)</td><td>\(12.5\)</td></tr> </table> Suppose she increases her weekly distance by the same percent each week. Write a model \(D(t)\) and predict her distance in week \(3\), rounded to the nearest tenth. Explain why adding \(2\) miles every week would not follow the model.

Hints

- Compare consecutive distances using both subtraction and division. - Which pattern would a fixed percent increase produce? - Use the model for the next week, then compare its increases in miles.

Solution

1. The initial distance is \(8\) miles. Consecutive ratios are \(\frac{10}{8}=\frac{12.5}{10}=1.25\). 2. The model is \(D(t)=8(1.25)^t\). Week \(3\) gives \(D(3)=15.625\approx15.6\) miles. 3. The increases are \(2\), then \(2.5\), then \(3.125\) miles, so a fixed addition of \(2\) would not match the percentage model.

Answer

\(D(t)=8(1.25)^t\); week \(3\): about \(15.6\) miles. The weekly amount added changes; the multiplier stays at \(1.25\).
Exponential growth and decay functions 28
A bacterial culture is modeled by \(N(t)=200(1.5)^t\), where \(t\) is measured in hours. a) By what factor does the population change over \(2\) hours, \(3\) hours, and \(5\) hours? b) Find the population after half an hour and after \(15\) minutes. Round to the nearest whole bacterium.

Hints

- How does the exponent change when the time increases by several hours? - Keep the time unit in the exponent consistent. - Convert minutes to hours before using the model. - Think about how fractional exponents represent fractional time intervals.

Solution

1. Over \(\Delta t\) hours, the population changes by a factor of \((1.5)^{\Delta t}\). 2. The factors for \(2\), \(3\), and \(5\) hours are \((1.5)^2=2.25\), \((1.5)^3=3.375\), and \((1.5)^5=7.59375\). 3. Half an hour is \(0.5\) hour, so \(N(0.5)=200(1.5)^{0.5}\approx244.95\), or about \(245\) bacteria. 4. Fifteen minutes is \(0.25\) hour, so \(N(0.25)=200(1.5)^{0.25}\approx221.34\), or about \(221\) bacteria.

Answer

a) \(2\) hours: \(2.25\); \(3\) hours: \(3.375\); \(5\) hours: \(7.59375\) b) Half an hour: about \(245\) bacteria; \(15\) minutes: about \(221\) bacteria
Exponential growth and decay functions 29
Let \(f(x)=5(3)^x\). Determine the factor by which \(f(x)\) is multiplied when a) \(x\) increases by \(1\); b) \(x\) increases by \(4\); c) \(x\) decreases by \(2\); d) \(x\) increases by \(0.5\).

Hints

- Rewrite \(f(x+h)\) using exponent rules. - Identify the part equal to \(f(x)\) and the remaining multiplier. - Recall the meanings of negative and fractional exponents.

Solution

1. In general, \(f(x+h)=5(3)^{x+h}=5(3)^x(3)^h=f(x)(3)^h\). Therefore, the change factor is \(3^h\). 2. For \(h=1\), the factor is \(3^1=3\). 3. For \(h=4\), the factor is \(3^4=81\). 4. For \(h=-2\), the factor is \(3^{-2}=\frac{1}{9}\). 5. For \(h=0.5\), the factor is \(3^{0.5}=\sqrt{3}\approx1.732\).

Answer

a) \(3\) b) \(81\) c) \(\frac{1}{9}\) d) \(\sqrt{3}\approx1.732\)
Exponential growth and decay functions 30
Let \(f(x)=4^x\). 1. Find \(\frac{f(x+1)}{f(x)}\) for \(x=0\) and \(x=2\). What do you notice? 2. Find the factor by which \(f(x)\) is multiplied when \(x\) increases by \(s=1.5\). 3. Find the value of \(s\) for which the function value is multiplied by exactly \(32\). 4. Complete the statement for \(f(x)=a^x\): “When \(x\) increases by \(s\), the function value is multiplied by ...”

Hints

- Use exponent rules to simplify ratios of powers. - Interpret a rational exponent such as \(1.5\). - Rewrite both sides of \(4^s=32\) using the same base. - Replace \(x\) with \(x+s\) and factor the expression.

Solution

1. At \(x=0\), \(\frac{f(1)}{f(0)}=\frac{4}{1}=4\). At \(x=2\), \(\frac{f(3)}{f(2)}=\frac{64}{16}=4\). The ratio is constant and equals the base. 2. The factor is \(4^{1.5}=4\sqrt{4}=8\). 3. Solve \(4^s=32\). Since \(4=2^2\) and \(32=2^5\), \(2^{2s}=2^5\), so \(2s=5\) and \(s=2.5\). 4. In general, \(f(x+s)=a^{x+s}=a^xa^s=f(x)a^s\), so the factor is \(a^s\).

Answer

1. The ratio is \(4\) in both cases. 2. \(8\) 3. \(s=2.5\) 4. \(a^s\)
Exponential growth and decay functions 31
Find the positive base \(b\) of the exponential function \(y=b^x\) that passes through each point. a) \(P(4, 81)\) b) \(Q(-3, 0.125)\) c) \(R(1.5, 8)\)

Hints

- Substitute the coordinates into the function equation. - Think about which power or root undoes each exponent. - In the negative-exponent case, rewrite the decimal as a fraction.

Solution

1. At \(P(4, 81)\), \(b^4=81\), so \(b=81^{1/4}=3\) because \(b>0\). 2. At \(Q(-3, 0.125)\), \(b^{-3}=0.125=\frac{1}{8}\). Therefore \(b^3=8\), so \(b=2\). 3. At \(R(1.5, 8)\), \(b^{3/2}=8\), so \(b=8^{2/3}=4\).

Answer

a) \(b=3\) b) \(b=2\) c) \(b=4\)
Exponential growth and decay functions 32
An exponential function has the form \(f(x)=ca^x\), where \(a>0\). Find \(c\) and \(a\) in each case. a) The graph passes through \(A(0, 7)\) and \(B(2, 63)\). b) The graph passes through \(A(0, 200)\) and \(B(-1, 50)\).

Hints

- Use the point with \(x=0\) first. - Substitute the value of \(c\) before using the second point. - Solve the resulting equation involving a square or a reciprocal.

Solution

1. A point with \(x=0\) gives \(c\), because \(a^0=1\). 2. In a), \(c=7\). Using \((2, 63)\), \(63=7a^2\), so \(a^2=9\). Since \(a>0\), \(a=3\). 3. In b), \(c=200\). Using \((-1, 50)\), \(50=200a^{-1}=\frac{200}{a}\), so \(a=4\).

Answer

a) \(c=7\), \(a=3\) b) \(c=200\), \(a=4\)
Exponential growth and decay functions 33
Consider the exponential function \(f(x)=a^x\), where \(a>0\) and \(a\ne 1\). a) For one such function, the output is multiplied by \(4\) whenever \(x\) increases by \(2\). Find the base \(a\). b) Explain why \(a=1\) is excluded from the usual definition of an exponential growth or decay function. What would its graph look like? c) Explain why a negative base such as \(a=-2\) does not define a real-valued function for every real \(x\). Compare \(x=2\) and \(x=0.5\).

Hints

- Use the exponent laws to express the growth factor over an increase of \(2\). - Consider what happens when every power has base \(1\). - Interpret an exponent of \(0.5\) as a square root. - Ask whether that square root is real for a negative base.

Solution

1. For part a, increasing \(x\) by \(2\) multiplies the output by \(a^2\). Thus, \(a^2=4\). Since \(a>0\), \(a=2\). 2. For part b, if \(a=1\), then \(f(x)=1^x=1\) for every \(x\). This is a constant function, so its graph is the horizontal line \(y=1\), not a growth or decay curve. 3. For part c, \((-2)^2=4\) is real, but \((-2)^{0.5}=\sqrt{-2}\) is not a real number. Therefore, a negative base does not give a real-valued exponential function with domain \(\mathbb{R}\).

Answer

a) \(a=2\) b) \(a=1\) gives the constant function \(f(x)=1\), whose graph is the horizontal line \(y=1\). c) \((-2)^2\) is real, but \((-2)^{0.5}\) is not real, so the function cannot have all real numbers as its domain.
Exponential growth and decay functions 34
Let \(g(x)=(0.75)^x\). a) By what factor does the function value change when \(x\) increases by \(2\)? b) By what percent does the function value decrease when \(x\) increases by \(3\)? c) If \(y=g(x)\), express \(g(2x)\) in terms of \(y\).

Hints

- Raise the one-step factor to a power for several steps. - Convert a factor below \(1\) to a percent decrease. - Use the power-of-a-power rule for part c.

Solution

1. When \(x\) increases by \(2\), the function value is multiplied by \((0.75)^2=0.5625\). 2. When \(x\) increases by \(3\), the factor is \((0.75)^3=0.421875\). The percent decrease is \((1-0.421875)\cdot100\%=57.8125\%\). 3. Since \(g(2x)=(0.75)^{2x}=((0.75)^x)^2\) and \(y=(0.75)^x\), \(g(2x)=y^2\).

Answer

a) \(0.5625\) b) \(57.8125\%\) decrease c) \(g(2x)=y^2\)
Exponential growth and decay functions 35
Let \(f(x)=ca^x\), where \(a>0\) and \(c\ne0\). 1. Show that \(\frac{f(x-4)}{f(x)}\) is independent of \(x\), and express the constant ratio in terms of \(a\). 2. Find the ratio when \(a=0.8\).

Hints

- Substitute \(x-4\) into the function rule. - Apply the quotient rule for powers with the same base. - Simplify until the expression no longer contains \(x\). - Interpret a negative exponent as a reciprocal.

Solution

1. Substitute the function rule: \(\frac{f(x-4)}{f(x)}=\frac{ca^{x-4}}{ca^x}\). Canceling \(c\) and subtracting exponents gives \(a^{(x-4)-x}=a^{-4}=\frac{1}{a^4}\). This expression does not contain \(x\), so the ratio is constant. 2. For \(a=0.8\), the ratio is \((0.8)^{-4}=\frac{1}{(0.8)^4}=\frac{1}{0.4096}=2.44140625\).

Answer

1. \(\frac{f(x-4)}{f(x)}=a^{-4}=\frac{1}{a^4}\) 2. \(2.44140625\)
Exponential growth and decay functions 36
Analyze the exponential function \(f(x)=1.5(3)^x\). a) Find its y-intercept. b) Is it increasing or decreasing? Explain. c) Give its range. d) As \(x\) gets smaller and smaller, what happens to \(f(x)\)?

Hints

- Find the output when the input is \(0\). - How do powers of \(3\) change when the exponent increases? - Can a positive multiplier times \(3^x\) ever be zero or negative? - Try \(x=-1,-2,-3\) to see what happens far to the left.

Solution

1. \(f(0)=1.5(3)^0=1.5\), so the y-intercept is \((0, 1.5)\). 2. Since \(3>1\) and the starting multiplier is positive, the function increases as \(x\) increases. 3. Every power of \(3\) is positive, so the range is \((0, \infty)\). 4. For increasingly negative inputs, \(3^x\) becomes a smaller positive number. Thus \(f(x)\) gets closer and closer to \(0\), without reaching it.

Answer

a) \((0, 1.5)\) b) Increasing, because its base is greater than \(1\). c) \((0, \infty)\) d) It gets closer and closer to \(0\), staying positive.
Exponential growth and decay functions 37
Let \(h(x)=5(2)^{-3x}\). The input \(x\) is increased repeatedly by \(\frac{1}{3}\). Find the factor by which the function value is multiplied at each step.

Hints

- Substitute \(x+\frac{1}{3}\) into the exponent. - Distribute \(-3\) carefully. - Rewrite a negative exponent as a reciprocal. - Compare the new function expression with the original one.

Solution

1. Substitute the new input: \(h\left(x+\frac{1}{3}\right)=5(2)^{-3\left(x+\frac{1}{3}\right)}\). 2. Simplify the exponent: \(-3\left(x+\frac{1}{3}\right)=-3x-1\). 3. Therefore, \(h\left(x+\frac{1}{3}\right)=5(2)^{-3x}(2)^{-1}=h(x)\left(\frac{1}{2}\right)\). 4. The multiplication factor is \(\frac{1}{2}\).

Answer

\(\frac{1}{2}\)
Exponential growth and decay functions 38
The graph of an exponential function \(f(x)=ca^x\) passes through \(A(2, 18)\) and \(B(4, 162)\). Find the function, then determine the missing y-coordinates of \(C(0, y_C)\) and \(D(-1, y_D)\).

Hints

- Substitute both points to create two equations. - Divide the equations to eliminate one parameter. - The base of an exponential function is positive. - Substitute each requested x-coordinate into the completed function.

Solution

1. The points give \(18=ca^2\) and \(162=ca^4\). 2. Divide the equations: \(\frac{162}{18}=a^2\), so \(a^2=9\). Since \(a>0\), \(a=3\). 3. Substitute into \(18=ca^2\): \(18=9c\), so \(c=2\). Thus \(f(x)=2(3)^x\). 4. \(y_C=f(0)=2\), and \(y_D=f(-1)=2\cdot3^{-1}=\frac{2}{3}\).

Answer

\(f(x)=2(3)^x\) \(y_C=2\), \(y_D=\frac{2}{3}\)
Exponential growth and decay functions 39
The table shows values of a function \(g\): <table> <tr><td>\(x\)</td><td>\(-1\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td></tr> <tr><td>\(g(x)\)</td><td>\(12\)</td><td>\(6\)</td><td>\(3\)</td><td>\(1.5\)</td></tr> </table> a) Identify the function family and write an equation for \(g\). b) Describe the change in the outputs when \(x\) increases by \(1\). c) Find \(g(0.5)\). Round to the nearest hundredth. Explain why this input is allowed even though it is not in the table.

Hints

- Compare ratios of neighboring outputs rather than their differences. - Which table column shows the starting value for the model? - Think about what an exponent of \(0.5\) means.

Solution

1. Consecutive output ratios are \(\frac{6}{12}=\frac{3}{6}=\frac{1.5}{3}=0.5\), so this is an exponential pattern. 2. Since \(g(0)=6\), \(g(x)=6(0.5)^x\). 3. Each time the input increases by \(1\), the output is multiplied by \(0.5\). 4. \(g(0.5)=6(0.5)^{0.5}=6\sqrt{0.5}\approx4.24\). The exponential rule has a positive base, so it also gives a real output for fractional inputs.

Answer

a) Exponential: \(g(x)=6(0.5)^x\) b) Each output is half the preceding output. c) \(g(0.5)\approx4.24\); a positive base can be raised to a fractional exponent.
Exponential growth and decay functions 40
An exponential function has the form \(g(x)=b^x\). Its graph passes through \((0.75, 8)\). Find \(b\).

Hints

- Rewrite \(0.75\) as a fraction. - Raise both sides to the reciprocal of the exponent. - Use exponent rules to simplify the result.

Solution

1. Substitute the point: \(8=b^{0.75}=b^{3/4}\). 2. Raise both sides to the power \(\frac{4}{3}\): \(b=8^{4/3}\). 3. Since \(\sqrt[3]{8}=2\), \(b=2^4=16\).

Answer

\(b=16\)
Exponential growth and decay functions 41
Consider \(f(x)=4^x\) and \(g(x)=4.1^x\). For \(x>0\), which function grows faster? Justify your answer by comparing their growth factors and describing what happens to \(\frac{g(x)}{f(x)}\) as \(x\) increases.

Hints

- How can you compare changes over a one-unit interval? - Rewrite the ratio of the two powers as one power. - What does the new base tell you about the ratio as \(x\) increases?

Solution

1. For a one-unit increase in \(x\), \(f\) is multiplied by \(4\) and \(g\) is multiplied by \(4.1\). 2. Since \(4.1>4\), \(g\) has the greater growth factor. 3. \(\frac{g(x)}{f(x)}=\left(\frac{4.1}{4}\right)^x=(1.025)^x\), which increases because \(1.025>1\). 4. Thus \(g(x)\) grows faster and pulls farther ahead as \(x\) increases.

Answer

\(g(x)=4.1^x\) grows faster; its factor is \(4.1\), compared with \(4\), and \(\frac{g(x)}{f(x)}=(1.025)^x\) increases.
Exponential growth and decay functions 42
The figure shows four graphs, labeled \(p\), \(q\), \(r\), and \(s\), of functions of the form \(y=b^x\). Match each graph with one function. Justify your matches using growth or decay, steepness, or key points. \(A(x)=3^x\) \(B(x)=1.5^x\) \(C(x)=0.5^x\) \(D(x)=0.2^x\)
Figure for problem 532299

Hints

- Use the base to determine whether a graph increases or decreases. - All four graphs share one point on the y-axis. - Evaluate the functions at \(x=1\) or \(x=-1\). - Compare the growth rates of the two increasing graphs. - Compare how quickly the two decreasing graphs fall.

Solution

1. Graphs \(p\) and \(q\) are increasing, so they correspond to bases greater than \(1\): \(3\) and \(1.5\). Graphs \(r\) and \(s\) are decreasing, so they correspond to bases between \(0\) and \(1\): \(0.5\) and \(0.2\). 2. Since \(3^x\) grows faster than \(1.5^x\) for positive \(x\), graph \(p\) is \(A\) and graph \(q\) is \(B\). At \(x=1\), their values are \(3\) and \(1.5\), respectively. 3. Since \(0.2^x\) decreases faster than \(0.5^x\) for positive \(x\), graph \(s\) is \(D\) and graph \(r\) is \(C\). At \(x=-1\), their values are \(5\) and \(2\), respectively.

Answer

Graph \(p\): \(A(x)=3^x\); graph \(q\): \(B(x)=1.5^x\); graph \(r\): \(C(x)=0.5^x\); graph \(s\): \(D(x)=0.2^x\)
Exponential growth and decay functions 43
A biologist compares two bacterial cultures. Each starts with one million bacteria. Culture A doubles each hour, \(f(x)=2^x\), while Culture B is halved each hour, \(g(x)=0.5^x\). Population is measured in millions. The model uses \(x\ge 0\), while the graph shows both complete functions. a) Describe the two graphs and explain their symmetry. b) Find the population ratio \(\frac{f(x)}{g(x)}\) at \(x=3\). c) Show that \(f(x)g(x)\) is constant for every \(x\), and state its value.
Figure for problem 533501

Hints

- Where do both curves meet the y-axis? Compare their shapes to the left and right of it. - Evaluate both models at the time requested before comparing their values. - For the product, look for a way to combine two powers with the same exponent.

Solution

1. For \(x\ge0\), \(f\) increases and \(g\) decreases. Both pass through \((0, 1)\). The complete graphs are reflections across the y-axis because \(g(x)=f(-x)\). 2. At \(x=3\), \(f(3)=8\) and \(g(3)=0.125\). Their ratio is \(\frac{8}{0.125}=64\). 3. \(f(x)g(x)=2^x(0.5)^x=(2\cdot0.5)^x=1\).

Answer

a) For \(x\ge0\), \(f\) increases and \(g\) decreases. The complete graphs reflect across the y-axis. b) \(64\) c) \(f(x)g(x)=1\) for all real \(x\).
Exponential growth and decay functions 44
Find an equation of the form \(y=ab^x\) for each graph \(f\) and \(g\). Use the y-intercept and one other clearly readable point from each graph.
Figure for problem 533505

Hints

- The parameter \(a\) is the function value at \(x=0\). - Choose a second point with coordinates that can be read exactly. - Substitute that point into \(y=ab^x\) and solve for \(b\).

Solution

1. For \(f\), the y-intercept is \((0, 5)\), so \(a=5\). The point \((1, 6)\) gives \(6=5b\), so \(b=\frac{6}{5}=1.2\). Therefore, \(f(x)=5(1.2)^x\). 2. For \(g\), the y-intercept is \((0, 2)\), so \(a=2\). The point \((1, 1)\) gives \(1=2b\), so \(b=0.5\). Therefore, \(g(x)=2(0.5)^x\).

Answer

\(f(x)=5(1.2)^x\) \(g(x)=2(0.5)^x\)
Exponential growth and decay functions 45
The table shows the number of users of a new app from week \(0\) through week \(3\). <table> <tr><th>Week \(t\)</th><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><th>Users \(y\)</th><td>\(10\)</td><td>\(30\)</td><td>\(90\)</td><td>\(270\)</td></tr> </table> a) Show that the data represent exponential growth, and give the growth factor \(b\). b) Write a function in the form \(y=ab^t\). c) How many users does the model predict at the end of week \(4\)?

Hints

- Compare consecutive values using ratios. - The initial value is the value at \(t=0\). - Substitute \(t=4\) into the model.

Solution

1. The ratios of consecutive values are \(\frac{30}{10}=3\), \(\frac{90}{30}=3\), and \(\frac{270}{90}=3\). The constant ratio shows exponential growth with factor \(b=3\). 2. The initial value is \(a=10\), so the model is \(y=10(3)^t\). 3. At \(t=4\), \(y=10\cdot3^4=810\).

Answer

a) \(b=3\) b) \(y=10(3)^t\) c) \(810\) users
Exponential growth and decay functions 46
Analyze each statement and draw the requested conclusion. a) Suppose \(0.6^m<0.6^n\). What relationship must hold between \(m\) and \(n\)? b) For a positive base \(a\), suppose \(a^5>a^7\). What interval must contain \(a\)? c) Suppose \(2^k=0.2\). Find two consecutive integers between which \(k\) lies.

Hints

- Determine whether each exponential function is increasing or decreasing. - Compare the order of the exponents with the order of the powers. - Find nearby powers of \(2\) that bracket \(0.2\).

Solution

1. Since \(0<0.6<1\), the function \(0.6^x\) is decreasing. Therefore, \(0.6^m<0.6^n\) implies \(m>n\). 2. Because \(5<7\) but \(a^5>a^7\), the exponential function must be decreasing. Thus \(0<a<1\). 3. \(2^{-3}=0.125\) and \(2^{-2}=0.25\). Since \(0.125<0.2<0.25\) and \(2^x\) is increasing, \(-3<k<-2\).

Answer

a) \(m>n\) b) \(0<a<1\) c) \(-3<k<-2\)
Exponential growth and decay functions 47
A frozen cake at \(-18\,^\circ\text{C}\) is placed in a kitchen at \(22\,^\circ\text{C}\). The temperature difference between the cake and the room decreases exponentially. After \(10\) minutes, the difference is \(32\,^\circ\text{C}\). a) Find the initial temperature difference. b) Find the decay factor per minute for the temperature difference, and write a function for the difference after \(t\) minutes. Round the per-minute decay factor to four decimal places. c) Find the cake's temperature after \(30\) minutes. d) Make a table of the temperature difference after \(0\), \(20\), \(40\), and \(60\) minutes. Round each table value to the nearest hundredth.

Hints

- Account for the negative initial temperature when finding the difference. - Use the value after \(10\) minutes to find either a 10-minute factor or a 1-minute factor. - The cake approaches room temperature from below. - For the table, use the fact that the difference is multiplied by \(0.8\) every \(10\) minutes.

Solution

1. The initial difference is \(22-(-18)=40\,^\circ\text{C}\). 2. If \(a\) is the decay factor per minute, then \(40a^{10}=32\). Thus \(a^{10}=0.8\), so \(a=0.8^{1/10}\approx0.9779\). The exact model is \(D(t)=40\left(0.8^{1/10}\right)^t\). 3. After \(30\) minutes, \(D(30)=40\cdot(0.8)^3=20.48\,^\circ\text{C}\). Since the cake is below room temperature, its temperature is \(22-20.48=1.52\,^\circ\text{C}\). 4. \(D(0)=40.00\), \(D(20)=25.60\), \(D(40)=16.384\approx16.38\), and \(D(60)=10.48576\approx10.49\), all in degrees Celsius.

Answer

a) \(40\,^\circ\text{C}\) b) \(a=0.8^{1/10}\approx0.9779\); \(D(t)=40\left(0.8^{1/10}\right)^t\) c) \(1.52\,^\circ\text{C}\) d) <table> <tr> <th>Time (min)</th> <th>Temperature difference (°C)</th> </tr> <tr> <td>\(0\)</td> <td>\(40.00\)</td> </tr> <tr> <td>\(20\)</td> <td>\(25.60\)</td> </tr> <tr> <td>\(40\)</td> <td>\(16.38\)</td> </tr> <tr> <td>\(60\)</td> <td>\(10.49\)</td> </tr> </table>
Exponential growth and decay functions 48
Find the parameters \(a\) and \(c\) in \(y=ca^x\) so that the graph passes through \(P(1, 4)\) and \(Q(-2, 32)\). Then find \(y_R\) and \(y_S\) for the points \(R(2, y_R)\) and \(S(-3, y_S)\).

Hints

- Write one equation for each given point. - Isolate one variable and substitute it into the other equation. - Apply the negative-exponent rule. - Use the completed function to find the two requested outputs.

Solution

1. The two points give \(4=ca\) and \(32=ca^{-2}\). 2. From \(4=ca\), \(c=\frac{4}{a}\). Substitute into the second equation: \(32=\frac{4}{a}a^{-2}=4a^{-3}=\frac{4}{a^3}\). 3. Thus \(a^3=\frac{1}{8}\), so \(a=0.5\). Then \(c=\frac{4}{0.5}=8\). The function is \(y=8(0.5)^x\). 4. At \(x=2\), \(y_R=8\cdot(0.5)^2=2\). At \(x=-3\), \(y_S=8\cdot(0.5)^{-3}=64\).

Answer

\(a=0.5\), \(c=8\) \(y_R=2\), \(y_S=64\)
Exponential growth and decay functions 49
Find the equation of an exponential function \(f(x)=ab^x\) whose graph passes through \((-1, 12)\) and \((1.5, 0.375)\).

Hints

- Substitute both points into the general model. - Solve one equation for \(a\). - Combine powers of \(b\) using exponent rules. - Raise both sides to the reciprocal of the remaining exponent.

Solution

1. Substitution gives \(12=ab^{-1}\) and \(0.375=ab^{1.5}\). 2. From the first equation, \(a=12b\). Substitute into the second: \(0.375=12b^{2.5}\). 3. Thus, \(b^{2.5}=0.03125\), so \(b=(0.03125)^{1/2.5}=0.25\). 4. Then \(a=12(0.25)=3\), so \(f(x)=3(0.25)^x\).

Answer

\(f(x)=3(0.25)^x\)

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