Skip to main content

Algebra and Precalculus A Practical Guide

Cameron Geisler

Section 13.2 Intro to Exponential Growth and Decay

Exponential functions can be used to model amounts that change over time.
🔗

Example 13.2.1. Population Growth.

A small town has a population of 12000 people. Each year, the population increases by 10%.
🔗
  1. What will be the population of the town after 1 year? 2 years? 3 years?
    🔗
    🔗
  2. What will be the population after \(t\) years?
    🔗
    🔗
“Increase by 10%” means that it keeps all of the population it has, and adds 10% more of that number.
🔗
  • 100% of the population is the amount it already has
    🔗
    🔗
  • 10% more means you are at 110% of the original.
    🔗
    🔗
As a decimal, \(110\%=1.10\text{.}\) So, increase by 10% means multiply by 1.10.
🔗
  • At the start (year 0), the population is 12000.
    🔗
    🔗
  • After 1 year, the population is \(12000 \times 1.10 = 13200\) people.
    🔗
    🔗
  • After 2 years, the population is \(12000 \times 1.10 \times 1.10 = 12000(1.10)^2 = 14520\) people.
    🔗
    🔗
  • After 3 years, the population is \(12000 \times 1.10 \times 1.10 \times 1.10 = 12000(1.10)^3 = 15972\) people.
    🔗
    🔗
Notice the pattern, which is that we multiply the initial population of 12000 by 1.10 for each year.
🔗
In general, after \(t\) years, we are multiplying by 1.10 precisely \(t\) times. So, the general formula for population, we’ll call it \(P(t)\text{,}\) is given by,
🔗
\begin{gather*} P(t)=12000(1.10)^t \end{gather*}
🔗

Example 13.2.2. Smartphone Depreciation.

Suppose you buy a new smartphone for $1400. Each year, its value drops by 35%.
🔗
  1. How much will it be worth after 1 year? 2 years? 3 years?
    🔗
    🔗
  2. How much will it be worth after \(t\) years?
    🔗
    🔗
“Drop by 35%” means you keep 65% of the value each year, because you start with 100% of its value, take away 35%, and are left with 65%. As a decimal, \(65\%=0.65\text{,}\) so decreasing by 35% each year means multiply by 0.65 each year.
🔗
\begin{align*} A(0) \amp = 1400\\ A(1) \amp = 1400(0.65)\\ A(2) \amp = 1400(0.65)^2\\ A(3) \amp = 1400(0.65)^3 \end{align*}
In general,
🔗
\begin{gather*} A(t)=1000(0.65)^t \end{gather*}
🔗
In general, exponential change has this form,
🔗
\begin{gather*} \text{amount after $t$ steps} = (\text{starting amount}) \times (\text{multiplier})^t \end{gather*}

Subsection 13.2.1 Exponential Growth and Decay Models

The exponential growth and exponential decay models are given by,
🔗
\begin{align*} \boxed{A = A_0 \cdot b^t \qquad \begin{cases} A \amp \rightarrow \text{amount after } t \text{ time steps} \\ A_0 \amp \rightarrow \text{initial amount (at } t = 0\text{)} \\ b \amp \rightarrow\text{growth factor for each time step} \\ t \amp \rightarrow \text{number of time steps} \end{cases}} \end{align*}
The base \(b\) is called the growth factor, because it controls how the amount changes from one step to the next.
🔗
It can also be thought of in terms of the percentage change, as \(1+r\text{,}\)
🔗
\begin{align*} \boxed{A = A_0 \cdot (1+r)^t \qquad \begin{cases} A \amp \rightarrow\text{amount after } t \text{ time steps} \\ A_0 \amp \rightarrow\text{initial amount (at } t = 0\text{)} \\ r \amp \rightarrow\text{percentage change (as a decimal)} \\ t \amp \rightarrow\text{number of time steps} \end{cases}} \end{align*}
Here, \(r\) represents the percentage change (as a decimal),
🔗
  • If \(r \gt 0\text{,}\) then it is exponential growth, and \(r\) is the growth rate.
    🔗
    🔗
  • If \(r \lt 0\text{,}\) then it is exponential decay, and \(r\) is the decay rate.
    🔗
    🔗
In simple terms,
🔗
\begin{gather*} \boxed{\text{New amount} = \text{Starting amount} \cdot (1 + \text{rate})^{\text{number of steps}}} \end{gather*}
Exponential functions have applications for,
🔗
  • Population growth (biology), of animals, people, or bacteria.
    🔗
    🔗
  • Growth of an investment (economics)
    🔗
    🔗
  • Decay of a radioactive substance (chemistry)
    🔗
    🔗
In general, anything that changes by a percentage can be modelled by exponential functions.
🔗

Exercise Group 13.2.1. Finding Growth Factors.

🔗

Subsection 13.2.2 Examples

Checkpoint 13.2.3. Bacteria Population Growth.

A microbiologist cultures a type of bacteria in a petri dish. The population \(B\) (in millions of bacteria) after \(t\) hours is modeled by \(B(t)=12(1.35)^t\text{.}\)
🔗
(a)
What is the initial bacteria population?
🔗
Answer.
\(B(0)=12\)
🔗
🔗
(c)
Find the bacteria population after 2 hours.
🔗
Answer.
\(B(2)=12(1.35)^2\approx 21.9\)
🔗
🔗
(d)
Find the bacteria population after 5 hours.
🔗
Answer.
\(B(5)=12(1.35)^5\approx 53.6\)
🔗
🔗
(e)
Find the bacteria population after 8 hours.
🔗
Answer.
\(B(8)=12(1.35)^8\approx 133.1\)
🔗
🔗
(f)
Sketch a graph of the bacteria population over time using the previous parts.
🔗
Answer.
An increasing exponential curve with \(y\)-intercept \((0,12)\text{,}\) passing through approximately \((2,21.9)\text{,}\) \((5,53.6)\text{,}\) and \((8,133.1)\text{.}\)
🔗
🔗
🔗
(g)
Estimate how long it will take for the bacteria population to reach 200 million, using a graphing calculator, or guess and test.
🔗
Answer.
\(t\approx 9.4\)
🔗
🔗
(h)
What is the domain and range of the function in this context?
🔗
Answer.
Domain is \(t \geq 0\text{,}\) range is \(B \geq 12\text{.}\)
🔗
🔗
🔗
🔗

Checkpoint 13.2.4. Insect Colony Population Growth.

An ecologist monitors a colony of insects in a controlled habitat. The population \(N\) (in thousands of insects) after \(t\) days is modeled by \(N(t)=8(1.22)^t\text{.}\)
🔗
(a)
What is the initial insect population?
🔗
Answer.
\(N(0)=8\)
🔗
🔗
(c)
Find the insect population after 4 days.
🔗
Answer.
\(N(4)=8(1.22)^4\approx 17.8\)
🔗
🔗
(d)
Find the insect population after 8 days.
🔗
Answer.
\(N(8)=8(1.22)^8\approx 39.3\)
🔗
🔗
(e)
Sketch a graph of the insect population over time using the previous parts.
🔗
Answer.
An increasing exponential curve with \(y\)-intercept \((0,8)\text{,}\) passing through approximately \((4,17.8)\) and \((8,39.3)\text{.}\)
🔗
🔗
🔗
(f)
Estimate how long it will take for the insect population to reach 80 thousand insects, using a graphing calculator, or guess and test.
🔗
Answer.
\(t\approx 11.6\)
🔗
🔗
🔗

Checkpoint 13.2.5. Coffee and Caffeine.

A cup of coffee contains 160 mg of caffeine. Caffeine in the body decreases with exponential decay, where the amount of caffeine in the body decreases by 13% per hour.
🔗
(a)
Write an equation that gives the amount of caffeine \(A\) after \(t\) hours of drinking a cup of coffee.
🔗
Answer.
\(A(t)=160(0.87)^t\)
🔗
🔗
(b)
How much caffeine remains in the body after 2 hours? After 5 hours?
🔗
Answer.
\(A(2)\approx 121.1\)\(A(5)\approx 79.7\)
🔗
🔗
(c)
Research shows that by bedtime, you should have at most 30 mg of caffeine in your body (to avoid it negatively affecting your sleep quality). How long will it take to reach 30 mg?
🔗
Answer.
\(t\approx 12.0\)
🔗
🔗
(d)
Suppose that instead, you drink 2 cups of coffee. Estimate how long it will take for your body to have only 30 mg left, using a graphing calculator or guess and test.
🔗
Answer.
\(t\approx 17.0\)
🔗
🔗
🔗

Checkpoint 13.2.6. Flu Virus Spreading.

A flu virus spreads in a school. The number of people \(N\) infected after \(t\) days is given by \(N=4\cdot 1.332^t\text{.}\)
🔗
(a)
What is the initial number of people infected?
🔗
Answer.
\(N(0)=4\)
🔗
🔗
(b)
What is the growth rate of people infected per day?
🔗
Answer.
🔗
🔗
(c)
Find the number of people who were infected after 16 days.
🔗
Answer.
\(N(16)=4\cdot1.332^16 \approx 393\)
🔗
🔗
(d)
There are 1200 people in the school. Estimate the time \(t\) it will take for everybody in the school to catch the flu, using a graphing calculator, or guess and test.
🔗
Answer.
\(t\approx 19.9\)
🔗
🔗
🔗

Checkpoint 13.2.7. Depreciation of a Car.

A car was purchased for $15,500. The car decreases in value by 15% each year.
🔗
(a)
Write an equation that would give the value \(V\) of the car \(t\) years after it was purchased.
🔗
Answer.
\(V=15500(0.85)^t\)
🔗
🔗
(b)
How many years would the car be worth one quarter of its original value?
🔗
Hint.
\(V=15500(0.85)^t\text{,}\) find \(t\) when \(V=3875\)
🔗
🔗
Answer.
\(t\approx 8.5\)
🔗
🔗
(c)
How many years would it take the car to be worth $2500?
🔗
Hint.
\(V=15500(0.85)^t\text{,}\) find \(t\) when \(V=2500\)
🔗
🔗
Answer.
\(t\approx 11.2\)
🔗
🔗
🔗

Checkpoint 13.2.8. Availability of Light Under Water.

As a scuba diver, the deeper you dive, the less light available to you (because the more light is absorbed by the water above you). Suppose that the amount of light available decreases by 12% for every 10 m that you descend.
🔗
(a)
Write the exponential function that relates the amount, \(L\text{,}\) as a percent expressed as a decimal, of light available to the depth, \(d\text{,}\) in 10-m increments.
🔗
Hint.
Each 10 m multiplies the light by \(1-\frac{12}{100}=0.88\text{,}\) and \(L(0)=1\)
🔗
🔗
Answer.
\(L(d)=0.88^d\)
🔗
🔗
(b)
Sketch a graph of the function.
🔗
Answer.
The graph is a decreasing exponential curve, with a \(y\)-intercept of 1.
🔗
🔗
🔗
(c)
What are the domain and range of the function in this context?
🔗
Answer.
Domain is \(d\geq 0\text{,}\) range is \(0 \lt L\leq 1\text{.}\)
🔗
🔗
🔗
(d)
What percent of light will reach you if you dive to a depth of 25 m?
🔗
Hint.
25 m is \(\frac{25}{10}=2.5\) increments of 10 m, so \(d=2.5\)
🔗
🔗
Answer.
\(L(2.5)=0.88^{2.5}\approx 0.73\text{,}\)
🔗
🔗
🔗
🔗

Subsection 13.2.3 Compound Interest

Another application of exponential growth is compound interest, where your interest also earns interest.
🔗
When a bank pays interest, it adds some money to your account. If you leave that interest in the account, then the next time interest is calculated, it’s calculated on a bigger amount than before. So the amount grows faster over time because it’s not just the original money earning interest, but the added interest also starts earning interest. This is precisely exponential growth.
🔗

Example 13.2.9.

Suppose that you deposit $2500 and your account earns 4% per year. After 1 year, the bank will pay you,
🔗
\begin{gather*} \$2500 \times 0.04 = \$100 \end{gather*}
This means that after 1 year, your new total is,
🔗
\begin{align*} \text{new amount} & = \text{old amount} + \text{interest}\\ & = 2500 + 100\\ & = \$2600 \end{align*}
Just like other examples of exponential growth, we can think of adding this interest like a growth factor. In this case, it is \(1+0.04=1.04\text{,}\) because to increase by 4% each year, we have to multiply by 1.04. Then, the amount of money \(A\) after \(t\) years is given by,
🔗
\begin{gather*} A=2500(1.04)^t \end{gather*}
🔗
In general,
🔗
\begin{align*} \boxed{A = P(1+r)^t \quad \begin{cases} A & \rightarrow \text{amount of money after } t \text{ years} \\ P & \rightarrow \text{principal (initial deposit at } t=0\text{)} \\ r & \rightarrow \text{annual interest rate (as a decimal)} \\ t & \rightarrow \text{number of years} \end{cases}} \end{align*}

Checkpoint 13.2.10. Compound Interest Calculation.

You deposit $2450 into a savings account that earns 3.2% interest per year, compounded annually. Write an equation for the amount \(A\) after \(t\) years, and find the amount after 4 years.
🔗
Hint.
Use \(A=P(1+r)^t\) with \(P=2450\) and \(r=0.032\)
🔗
🔗
Answer.
\(A=2450(1.032)^t\)\(A(4)\approx \$2778.98\)
🔗
🔗

Checkpoint 13.2.11. Scholarship Fund.

A scholarship fund has $3175 invested at 2.7% interest per year, compounded annually. How much will be in the fund after 2 years and after 7 years?
🔗
Hint.
Use \(A=3175(1.027)^t\) and evaluate at \(t=2\) and \(t=7\)
🔗
🔗
Answer.
\(A(2)\approx \$3348.76\)\(A(7)\approx \$3825.93\)
🔗
🔗

Checkpoint 13.2.12. Total Interest Earned.

You invest $4325 at 4.2% interest per year, compounded annually, for 6 years. How much interest did you earn in total?
🔗
Hint.
First find \(A=4325(1.042)^6\text{,}\) then subtract your principal
🔗
🔗
Answer.
\(\$1210.95\)
🔗
🔗

Checkpoint 13.2.13. Yearly Interest.

You deposit $5200 into an account that earns 4.5% interest per year, compounded annually. How much interest is earned in the first year, and how much interest is earned in the fifth year only?
🔗
Hint.
\(A(t)=5200(1.045)^t\text{,}\) first-year interest is \(A(1)-A(0)\text{,}\) and fifth-year interest is \(A(5)-A(4)\)
🔗
🔗
Answer.
\(\approx \$234.00\text{,}\)\(\approx \$279.05\)
🔗
🔗

Checkpoint 13.2.14. Future Value Goal.

You want to have $12000 in 9 years. The account pays 3.8% interest per year, compounded annually. What single deposit do you need to make today?
🔗
Hint.
Use \(12000=P(1+0.038)^9\) and solve for \(P\)
🔗
🔗
Answer.
\(P\approx \$8578.38\)
🔗
🔗

Checkpoint 13.2.15. Saving for Laptop.

A student wants $5000 for a laptop in 4 years. A savings account pays 5.5% interest per year, compounded annually. What single deposit is needed today?
🔗
Hint.
Use \(5000=P(1.055)^4\) and solve \(P=\frac{5000}{(1.055)^4}\)
🔗
🔗
Answer.
\(P\approx \$4036.08\)
🔗
🔗

Checkpoint 13.2.16. Time to Reach Goal.

You deposit $1500 into an account that earns 6% interest per year, compounded annually. Estimate how long will it take for the account to reach $2500, using a graphing calculator or guess and test.
🔗
Hint.
\(A=1500(1.06)^t\text{,}\) find \(t\) when \(A=2500\)
🔗
🔗
Answer.
\(t\approx 8.8\)
🔗
🔗

Checkpoint 13.2.17. Doubling Time.

You deposit $2400 into an account that earns 7% interest per year, compounded annually. About how long will it take for the money to double?
🔗
Hint.
Set up \(4800=2400(1.07)^t\)
🔗
🔗
Answer.
\(t\approx 10.2\)
🔗
🔗

Checkpoint 13.2.18. Finding Interest Rate.

An investment grows from $5000 to $5800 in 3 years with compound interest compounded annually. Estimate the annual interest rate \(r\text{.}\)
🔗
Hint.
Set up \(5800=5000(1+r)^3\) and use guess and test for \(r\text{.}\)
🔗
🔗
Answer.
\(r\approx 0.051\text{,}\)\(5.1\%\)
🔗
🔗

Checkpoint 13.2.19. Finding Initial Deposit.

An account has $8420 after 6 years at 3.9% interest per year, compounded annually. What was the initial deposit?
🔗
Hint.
Use \(8420=P(1.039)^6\) and solve for \(P\)
🔗
🔗
Answer.
\(P\approx \$6692.97\)
🔗
🔗

Checkpoint 13.2.20. Investment Growth Time.

You invest $6800 at 4.2% interest per year, compounded annually. About how long will it take to reach $9000?
🔗
Hint.
\(A=6800(1.042)^t\text{,}\) find \(t\) when \(A=9000\)
🔗
🔗
Answer.
\(t\approx 6.8\)
🔗
🔗
🔗

Subsection 13.2.4 More Examples

Checkpoint 13.2.21. Grasshopper Plague.

A scientist monitoring a grasshopper plague notices that the area affected by the grasshoppers is given by \(A(n)=1000(1.15)^n\) hectares, where \(n\) is the number of weeks after the initial observation.
🔗
(a)
What is the original affected area?
🔗
Answer.
\(A(0)=1000\)
🔗
🔗
(b)
What is the growth rate per week for the affected area?
🔗
Answer.
🔗
🔗
(c)
Find the affected area after 5 weeks.
🔗
Answer.
\(A(5)\approx 2011.4\)
🔗
🔗
(d)
Find the affected area after 10 weeks.
🔗
Answer.
\(A(10)\approx 4045.6\)
🔗
🔗
(e)
Sketch a graph of the affected area over time.
🔗
Answer.
An increasing exponential curve with \(y\)-intercept \((0,1000)\text{,}\) passing through approximately \((5,2011.4)\) and \((10,4045.6)\text{.}\)
🔗
🔗
🔗
(f)
Use a graphing calculator to find how long it will take for the affected area to reach 8000 hectares.
🔗
Answer.
\(n\approx 14.9\)
🔗
🔗
🔗

Checkpoint 13.2.22. Bacteria Weight.

The weight \(W\) of bacteria in a culture \(t\) hours after establishment is given by \(W(t)=100(1.07)^t\) grams.
🔗
(a)
Find the initial weight.
🔗
Answer.
\(W(0)=100\)
🔗
🔗
(b)
Find the weight after 4 hours.
🔗
Answer.
\(W(4)\approx 131.1\)
🔗
🔗
(c)
Find the weight after 10 hours.
🔗
Answer.
\(W(10)\approx 196.7\)
🔗
🔗
(d)
Find the weight after 24 hours.
🔗
Answer.
\(W(24)\approx 507.2\)
🔗
🔗
(e)
Sketch a graph of the bacteria weight over time using the results of the previous parts.
🔗
Answer.
An increasing exponential curve with \(y\)-intercept \((0,100)\) and points \((4,131.1)\text{,}\) \((10,196.7)\text{,}\) \((24,507.2)\text{.}\)
🔗
🔗
🔗
🔗

Checkpoint 13.2.23. Pygmy Possum Population.

A breeding program to ensure the survival of pygmy possums is established with an initial population of 50 (25 pairs). From a previous program, the expected population \(P\) in \(n\) years time is given by \(P(n)=P_0(1.23)^n\text{.}\)
🔗
(a)
What is the value of \(P_0\text{?}\)
🔗
Answer.
\(P_0=50\)
🔗
🔗
(b)
Find the expected population after 2 years.
🔗
Answer.
\(P(2)\approx 75.6\)
🔗
🔗
(c)
Find the expected population after 5 years.
🔗
Answer.
\(P(5)\approx 140.8\)
🔗
🔗
(d)
Find the expected population after 10 years.
🔗
Answer.
\(P(10)\approx 396.3\)
🔗
🔗
(e)
Sketch a graph of the population over time using the previous parts.
🔗
Answer.
An increasing exponential curve with \(y\)-intercept \((0,50)\) and points \((2,75.6)\text{,}\) \((5,140.8)\text{,}\) \((10,396.3)\text{.}\)
🔗
🔗
🔗
(f)
Find the approximate time needed for the population to reach 500, using a graphing calculator or guess and test.
🔗
Answer.
\(n\approx 11.1\)
🔗
🔗
🔗

Checkpoint 13.2.24. Bear Population.

A species of bear is introduced to a large island off Alaska where previously there were no bears. In 1998, 6 pairs of bears were introduced. It is expected that the population will increase according to \(B(t)=B_0(1.13)^t\) where \(t\) is the time, in years, since the introduction.
🔗
(b)
Find the expected bear population in 2018.
🔗
Answer.
\(t=20\text{,}\)\(B(20)\approx 138.3\)
🔗
🔗
(c)
Find the expected percentage increase in population from 2008 to 2018.
🔗
Hint.
In 2008, \(t=10\text{,}\) so \(B(10)\approx 40.7\) bears
🔗
🔗
Answer.
\(t=10\)\(t=20\text{,}\)\(239.5\%\)
🔗
🔗
(d)
Find how long approximately it will take for the population to reach 200, using a graphing calculator or guess and test.
🔗
Answer.
\(t\approx 23.0\)
🔗
🔗
🔗

Checkpoint 13.2.25. Light Intensity Depth.

The intensity of light is reduced by 2% for each metre that a diver descends below the surface of the water. At what depth is the intensity of light only 10% of that at the surface?
🔗
Hint.
\(I=I_0(0.98)^d\text{,}\) find \(d\) when \(I=0.1 I_0\)
🔗
🔗
Answer.
\(I=I_0(0.98)^d\text{,}\)\(d=\frac{\log{0.1}}{\log{0.98}}\approx 114\)
🔗
🔗

Checkpoint 13.2.26. World Population Growth.

Statistics indicate that the world population since 1995 has been growing at a rate of about 1.27% per year. United Nations records estimate that the world population in 2011 was approximately 7 billion. Assuming the same exponential growth rate, when will the population of the world be 9 billion?
🔗
Hint.
Use \(P=7(1.0127)^t\text{,}\) find \(t\) when \(P=9\text{,}\) where \(t\) is years after 2011
🔗
🔗
Answer.
\(t\approx 20\text{,}\)
🔗
🔗

Checkpoint 13.2.27. Krumbein Phi Scale.

The Krumbein phi scale is used in geology to classify sediments such as silt, sand, and gravel by particle size. The scale is modelled by the function \(D(\varphi)=2^{-\varphi}\text{,}\) where \(D\) is the diameter of the particle, in millimetres, and \(\varphi\) is the Krumbein scale value. Fine sand has a Krumbein scale value of approximately 3. Coarse gravel has a Krumbein scale value of approximately \(-5\text{.}\)
🔗
(a)
Why would a coarse material have a negative scale value?
🔗
Answer.
Coarser material has a larger diameter \(D\text{.}\) Since \(D(\varphi)=2^{-\varphi}\text{,}\) making \(\varphi\) negative makes \(-\varphi\) positive, which increases \(2^{-\varphi}\) and therefore increases the diameter.
🔗
🔗
🔗
(b)
How does the diameter of fine sand compare to the diameter of coarse gravel?
🔗
Answer.
\(D(3)=2^{-3}=\frac{1}{8}\)\(D(-5)=2^{-(-5)}=2^5=32\)\(\frac{D(-5)}{D(3)}=\frac{32}{\frac{1}{8}}=256\text{,}\)
🔗
🔗
🔗
🔗
🔗
PrevTopNext