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Applications of Exponential Functions

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Using Exponential Growth and Decay Models
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The half-life of an isotope of barium is about 10 years. The half-life of a substance is the amount of time it takes for half of that substance to decay. If a nuclear scientist starts with 200 grams of barium, how many grams will remain after 100 years?


When a real-life quantity increases by a percentage over a period of time, the final amount can be modeled by the equation: A=P(1+r)^t , where A is the final amount, P is the initial amount, r is the rate (or percentage), and t is the time (in years). 1+r is known as the growth factor .

Conversely, a real-life quantity can decrease by a percentage over a period of time. The final amount can be modeled by the equation: A=P(1-r)^t , where 1-r is the decay factor .

Example A

The population of Coleman, Texas grows at a 2% rate annually. If the population in 2000 was 5981, what was the population is 2010? Round up to the nearest person.

Solution: First, set up an equation using the growth factor. r=0.02, t=10, and P=5981 .

A & =5981(1+0.02)^{10} \\& =5981(1.02)^{10} \\& =7291 \ \text{people}

Example B

You deposit $1000 into a savings account that pays 2.5% annual interest. Find the balance after 3 years if the interest rate is compounded a) annually, b) monthly, c) daily.

Solution: For part a, we will use A=1000(1.025)^3=1008.18 , as we would expect from Example A.

But, to determine the amount if it is compounded in amounts other than yearly, we need to alter the equation. For compound interest, the equation is A=P \left(1+ \frac{r}{n}\right)^{nt} , where n is the number of times the interest is compounded within a year. For part b, n=12 .

A & =1000 \left(1+ \frac{0.025}{12}\right)^{12 \cdot 3} \\& =1000(1.002)^{36} \\& =1077.80

In part c, n=365 .

A & =1000 \left(1+ \frac{0.025}{365}\right)^{365 \cdot 3} \\& =1000(1.000068)^{1095} \\& =1077.88

Example C

You buy a new car for $35,000. If the value of the car decreases by 12% each year, what will the value of the car be in 5 years?

Solution: This is a decay function because the value decreases .

A & =35000 (1-0.12)^5 \\& =35000(0.88)^{5} \\& =18470.62

The car would be worth $18,470.62 after five years.

Intro Problem Revisit This is an example of exponential decay, so we can once again use the exponential form f(x)=a \cdot b^{x-h}+k . In this case, a = 200, the starting amount; b is 1/2, the rate of decay; x-h = 100/10 = 10, and k = 0.

P = 200 \cdot {\frac{1}{2}}^{10}\\= 200 \cdot {\frac{1}{1024}} = 0.195

Therefore, 0.195 grams of the barium still remain 100 years later.

Guided Practice

1. Tommy bought a truck 7 years ago that is now worth $12,348. If the value of his truck decreased 14% each year, how much did he buy it for? Round to the nearest dollar.

2. The Wetakayomoola credit card company charges an Annual Percentage Rate (APR) of 21.99%, compounded monthly. If you have a balance of $2000 on the card, what would the balance be after 4 years (assuming you do not make any payments)? If you pay $200 a month to the card, how long would it take you to pay it off? You may need to make a table to help you with the second question.

3. As the altitude increases, the atmospheric pressure (the pressure of the air around you) decreases. For every 1000 feet up, the atmospheric pressure decreases about 4%. The atmospheric pressure at sea level is 101.3. If you are on top of Hevenly Mountain at Lake Tahoe (elevation about 10,000 feet) what is the atmospheric pressure?


1. Tommy needs to use the formula A = P(1-r)^t and solve for P .

12348 &= P(1-0.14)^7 \\12348 &= P(0.86)^7 \qquad \qquad \text{Tommy's truck was originally} \ \$35,490. \\\frac{12348}{(0.86)^7} &= P \approx 35490

2. you need to use the formula A=P \left(1+ \frac{r}{n}\right)^{nt} , where n=12 because the interest is compounded monthly.

A & =2000 \left(1+ \frac{0.2199}{12}\right)^{12 \cdot 4} \\& =2000(1018325)^{48} \\& =4781.65

To determine how long it will take you to pay off the balance, you need to find how much interest is compounded in one month, subtract $200, and repeat. A table might be helpful. For each month after the first, we will use the equation, B=R \left(1+ \frac{0.2199}{12}\right)^{12 \cdot (\frac{1}{12})} = R(1.018325) , where B is the current balance and R is the remaining balance from the previous month. For example, in month 2, the balance (including interest) would be B=1800 \left(1+ \frac{0.2199}{12}\right)^{12 \cdot \left(\frac{1}{12}\right)}= 1800 \cdot 1.08325=1832.99 .

Month 1 2 3 4 5 6 7 8 9 10 11
Balance 2000 1832.99 1662.91 1489.72 1313.35 930.09 790.87 640.06 476.69 299.73 108.03
Payment 200 200.00 200.00 200.00 200.00 200.00 200.00 200.00 200.00 200.00 108.03
Remainder $1800 1632.99 1462.91 1289.72 913.35 730.09 590.87 440.06 276.69 99.73 0

It is going to take you 11 months to pay off the balance and you are going to pay 108.03 in interest, making your total payment $2108.03.

3. The equation will be A=101,325(1-0.04)^{100}=1709.39 . The decay factor is only raised to the power of 100 because for every 1000 feet the pressure decreased. Therefore, 10,000 \div 1000=100 . Atmospheric pressure is what you don’t feel when you are at a higher altitude and can make you feel light-headed. The picture below demonstrates the atmospheric pressure on a plastic bottle. The bottle was sealed at 14,000 feet elevation (1), and then the resulting pressure at 9,000 feet (2) and 1,000 feet (3). The lower the elevation, the higher the atmospheric pressure, thus the bottle was crushed at 1,000 feet.


Growth Factor
The amount, 1+r , an exponential function grows by. Populations and interest commonly use growth factors.
Decay Factor
The amount, 1-r , an exponential function decreases by. Populations, depreciated values, and radioactivity commonly use decay factors.
Compounded Interest
When an amount of money is charges a particular interest rate and that rate is collected yearly, monthly, quarterly, or even daily. It is compounded because after the first “collection” interest is taken on interest.

Explore More

Use an exponential growth or exponential decay function to model the following scenarios and answer the questions.

  1. Sonya’s salary increases at a rate of 4% per year. Her starting salary is $45,000. What is her annual salary, to the nearest $100, after 8 years of service?
  2. The value of Sam’s car depreciates at a rate of 8% per year. The initial value was $22,000. What will his car be worth after 12 years to the nearest dollar?
  3. Rebecca is training for a marathon. Her weekly long run is currently 5 miles. If she increase her mileage each week by 10%, will she complete a 20 mile training run within 15 weeks?
  4. An investment grows at a rate of 6% per year. How much, to the nearest $100, should Noel invest if he wants to have $100,000 at the end of 20 years?
  5. Charlie purchases a 7 year old used RV for $54,000. If the rate of depreciation was 13% per year during those 7 years, how much was the RV worth when it was new? Give your answer to the nearest one thousand dollars.
  6. The value of homes in a neighborhood increase in value an average of 3% per year. What will a home purchased for $180,000 be worth in 25 years to the nearest one thousand dollars?
  7. The population of a community is decreasing at a rate of 2% per year. The current population is 152,000. How many people lived in the town 5 years ago?
  8. The value of a particular piece of land worth $40,000 is increasing at a rate of 1.5% per year. Assuming the rate of appreciation continues, how long will the owner need to wait to sell the land if he hopes to get $50,000 for it? Give your answer to the nearest year.

For problems 9-15, use the formula for compound interest: A=P \left(1+ \frac{r}{n}\right)^{nt} .

  1. If $12,000 is invested at 4% annual interest compounded monthly, how much will the investment be worth in 10 years? Give your answer to the nearest dollar.
  2. If $8,000 is invested at 5% annual interest compounded semiannually, how much will the investment be worth in 6 years? Give your answer to the nearest dollar.
  3. If $20,000 is invested at 6% annual interested compounded quarterly, how much will the investment be worth in 12 years. Give your answer to the nearest dollar.
  4. If $5,000 is invested at 8% annual interest compounded quarterly, how much will the investment be worth in 15 years? Give your answer to the nearest dollar.
  5. How much of an initial investment is required to insure an accumulated amount of at least $25,000 at the end of 8 years at an annual interest rate of 3.75% compounded monthly? Give your answer to the nearest one hundred dollars.
  6. How much of an initial investment is required to insure an accumulated amount of at least $10,000 at the end of 5 years at an annual interest rate of 5% compounded quarterly? Give your answer to the nearest one hundred dollars.
  7. Your initial investment of $20,000 doubles after 10 years. If the bank compounds interest quarterly, what is your interest rate?

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