Lesson Explainer: Applications of Exponential Functions Mathematics

In this explainer, we will learn how to solve real-world problems involving exponential functions.

Recall that the basic exponential function is given by 𝑓(π‘₯)=𝑏, where the base 𝑏 is a positive number other than 1. The general form is 𝑓(π‘₯)=𝐴𝑏, and, in β€œmodeling” real phenomena, we have

  • Independent variable π‘₯ is usually time.
  • Since 𝑓(0)=𝐴𝑏=𝐴, the quantity 𝐴 is the initial value of what the function is measuring. The value at time π‘₯=0.
  • The base 𝑏 tells us something about the β€œrate” at which our quantity is changing with time. Growth will correspond to a base 𝑏>1, while decay will be when 𝑏<1.

Assuming a positive value for 𝐴, the graphs all look like this:

Note that 𝐴𝑏=𝑓(1) is the amount after one unit of time. Notice also how the quantity changes from 𝐴 to 𝐴𝑏, depending on how 𝑏 compares to 1. Since 𝑓(2)=𝐴𝑏=(𝐴𝑏)π‘οŠ¨, we see that the values at times π‘₯=1,2,3,… are in the geometric sequence 𝐴𝑏,𝐴𝑏,𝐴𝑏,β€¦οŠ¨οŠ© with common ratio 𝑏. Thus, β€œdoubling” every unit of time ↔𝑏=2 while β€œhalving” ↔𝑏=12.

Here is an example.

Example 1: Forming an Expression to Model Real-World Exponential Growth

The number of bacteria in a laboratory quadruples every hour. There were initially 200 bacteria. Write an expression for 𝐡(𝑑), the number of bacteria 𝑑 hours after the initial measurement.


This is an example of exponential growth, so 𝐡(𝑑)=𝐴𝑏 for some constants 𝐴,𝑏 where independent variable time 𝑑 is measured in hours. The quadrupling means 𝑏=4, and 𝐴=𝐡(0) being the initial amount gives 𝐴=200, so 𝐡(𝑑)=200ο€Ή4.

In many cases, the rate 𝑏 is given as a β€œpercentage rate.” Recall what this means. Suppose that the value of a car depreciates by 4.5% each year. This means that if the value was 𝐴 at the beginning of the year, then, at the end, the value is less by 4.5% of this amount and is π΄βˆ’(4.5%)𝐴=𝐴(1βˆ’4.5%)=𝐴1βˆ’4.5100=𝐴(1βˆ’0.045)=𝐴(0.955).

If 𝑉(𝑑)=𝐴𝑏 expresses this depreciation, then 𝑏=1βˆ’0.045=0.955,𝑉(𝑑)=𝐴0.955.

Note that this rate is less than 1, as expected of a decaying process. Using percentage rates is talking about rates relative to 1, so a growth rate of π‘Ÿ means a rate of 𝑏=1+π‘Ÿ.

Here is another example.

Example 2: Identifying the Initial Value and the Rate of Increase of an Exponential Growth Function

The value, 𝑉(𝑑) dollars, of a property 𝑑 years from now can be modeled by the function 𝑉(𝑑)=300000Γ—1.075.

  1. What is the property’s value now?
  2. What is the rate of increase in the property’s value?


  1. The property’s value now is 𝑉(0)=300000Γ—1.075=300000Γ—1=300000.
  2. In the form 𝑉(𝑑)=𝐴𝑏, we see that 𝐴=300000 and 𝑏=1.075. Writing 1.075=𝑏=1+π‘Ÿ and solving for π‘Ÿ gives a percentage rate of increase in value of π‘Ÿ=0.075=7.5100=7.5%. So, the property’s value is increasing at a rate of 7.5% every year.

Banks often accrue interest on a periodic basis. For example if a β€œnominal” annual rate of 5% is collected every three months. This means that, in a single year, it is gathered at the rate of 54% four times in that year. So, starting with a loan of 𝑃 dollars, the total debt at year end is 𝑃1+54%οˆο€Ό1+54%οˆο€Ό1+54%οˆο€Ό1+54%=𝑃1+54%.οŠͺ

Since 1+1.25%=1+0.0125=1.0125, this represents a debt of 𝑃1.0125ο…β‰ˆπ‘ƒ(1.05095),οŠͺ which is what would be levied if the annual rate was actually 5.095%, levied just the one time. More generally, a (nominal) rate of π‘Ÿ percent over 𝑛 periods per year starting with an amount 𝑃 will grow, after 𝑑 years, to 𝑃1+π‘Ÿπ‘›ο‡οŠο.

Example 3: Making Calculation with Compound Interest

If $800 is earning interest semiannually at 2% per annum, what is the amount after 𝑛 years?


The earnings are over 𝑛=2 periods at a nominal rate of π‘Ÿ=2% with 𝑃=800 dollars. The formula for the amount after a single year is 𝑃1+π‘Ÿπ‘›ο‡=800ο€½1+2%2=800ο€Ό1+0.022=800(1.01) after 𝑛 years800(1.01)=800(1.0201).

The effective (or annual) rate is 2.01%.

Since exponential growth/decay is modeled with exponential functions, we can solve for the independent variable (say time) using logarithms.

Example 4: Solving Real-World Problems Involving Exponential Growth

A microorganism reproduces by binary fission, where every hour each cell divides into two cells. Given that there are 24β€Žβ€‰β€Ž431 cells to begin with, determine how long it will take for there to be 97β€Žβ€‰β€Ž724 cells.


The number of cells 𝑃(𝑑), after 𝑑 hours, is growing by 𝑃(𝑑)=𝐴𝑏, where 𝐴=𝑃(0)=24431 and the rate 𝑏=2 because of the β€œdoubling.” We must solve for 𝑑: 97724=𝑃(𝑑)=24431ο€Ή2, which gives 9772424431=24=22=(4)=𝑑.log

So, it will take 2 hours for the cell population to reach 97β€Žβ€‰β€Ž724.

Example 5: Interpreting Parameters in Exponential Functions in a Real-World Context

The number of Ebola infections in West Africa at the start of an epidemic followed an exponential growth. It is given by 𝑁=π‘’οŠ¦οŽ–οŠ¦οŠ­οŠ«ο, with 𝑑 the number of days after the first infection.

  1. What does the coefficient 0.075 represent?
    1. It is the number of new infections per day.
    2. It is the time it takes for the number of infections to be multiplied by 𝑒.
    3. 10.075 is the time it takes for the number of infections to be multiplied by 𝑒.
    4. It is the percentage of the daily growth in the number of infections 7.5%.
    5. It is the number of days after the first infection.
  2. By rewriting the formula in the form 𝑁=𝑏, find the percentage of the daily growth in the number of infections. Give your answer to one decimal place.


  1. We will discuss each of these options in turn.
    Option (A) does not seem reasonable: an exponential growth will not have a constant rise in the number of infections. This answer would fit a linear growth.
    (B) sounds better, since exponential figures grow multiplicatively. But this is not right. In a β€œdoubling” formula, we get the term 2 which says that time 𝑇 is when the quantity is multiplied by 2. This would be true if the formula read π‘’ο‘‰οŽŸο’οŽŸοŽ¦οŽ€ instead.
    (D) is saying that the daily rate of increase of the infected is 7.5%. This would correspond to a formula involving a term like (1.075).
    (E) is not very meaningful.
    (C) is correct in the same way that (B) was incorrect: setting 𝑑=10.075 in the formula gives 𝑁=𝑒, and when 𝑑=𝑛0.075, we get 𝑁=π‘’οŠ. Each interval of this amount of time multiplies the infections by 𝑒.
    The right answer is option (C).
  2. We rewrite as follows:𝑁=𝑒=𝑒=1.0779.οŠ¦οŽ–οŠ¦οŠ­οŠ«οοŠ¦οŽ–οŠ¦οŠ­οŠ«οο
    When the base is written as 𝑏=1+π‘Ÿ, the percentage rate of growth is exactly π‘βˆ’1 which is 0.0779=7.79%.

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