For Students

For Parents and Teachers

 

Home > Quizzes and Tutorials > Logarithms and Exponents > Real-World Applications of Exponential Functions

Real-World Applications of Exponential Functions

Before beginning this tutorial, it may be helpful to review the tutorial on Properties of Logarithms.

Now we can talk about exponential functions.  They appear in more places than you might think!

Example 1

Populations

  1. A tree frog population doubles every three weeks.  Suppose that currently, there are 10 tree frogs in your back yard.  How many tree frogs will there be in 6 months, assuming that there are four weeks each month?
  1. How long will it take this population to be 10,240?

Solutions

  1. We’ll first have to figure out how many times this population will double in 4 months.  With four weeks each month, the six months will be equal to 24 weeks.  Since the population will double every three weeks, it will double 8  times in 24 weeks.  Let’s look at the table below:

 

Number of     Number of                     
Weeks            doubling periods                  Population

0                                  0                                  10
3                                  1                                  20  = 10 x 2, or 10 x 21
6                                  2                                  40  = 10 x 2 x 2, or 10 x 22
9                                  3                                  80  = 10 x 2 x 2 x 2, or 10 x 23

 After 24 weeks, the population will be 10 x 28, or 2560 tree frogs!
           

  1. If you look at the table above, you will notice that you could let n be the number of weeks.  How would you get the number of doubling periods from n? 

 

That’s right, you divide n by 3!  So after n weeks, the population would be P = 10 x 2(n/3).

So we need to solve the equation 10,240 = 10 x 2(n/3).  If you divide both sides by 10, you’ll have 1024 = 2(n/3).

With some thought, you can recall that 1024 = 210.  Watch what results:
1024 = 2(n/3)
210 = 2(n/3)

So how can you find n? 

Think hard!  Both sides of the equation are written as powers of 2, so what has to be true?  Yes, the exponents, have to be the same!

So now we know that 10 = n/3, and n = 30.  That means that after 30 weeks, the population will be 10,240.

Example 2

Bacterial Decay

Currently, 80,000 bacteria are present in a culture. When an antibiotic is added to the culture, the number of bacteria is reduced by half every 3 hours. How many bacteria are left after a day? When will fewer than 1000 bacteria be present?

Solutions

After 3 hours, 80,000(1/2), or 40,000 bacteria are present.  A day would represent how many of these halving periods?  In other words, how many 3-hour time periods are there in a day?  Well, since a day has 24 hours, that would make 8 of these halving periods.

Number of     Number of                
hours              halving periods        Number of bacteria

0                      0                                  80,000
3                      1                                  40,000 = 80,000(1/2)
6                      2                                  20,000 = 80,000(1/2)(1/2), or 80,000(1/2)2
24                    8                                 ?

We’ll need to compute 80,000(1/2)8, which is 312.5.  Since we don’t really have half of a bacteria, we could round this answer up to about 313 bacteria after a day. 

Now when will fewer than 1000 bacteria be present?  We need to write an equation to represent this situation.  From the table above, we can see that the number of halving periods is the exponent placed on (1/2).  This means that we need to figure out the solution to the equation 80,000(1/2)(n/3) < 1000. 

Dividing both sides of the inequality by 80,000 gives (1/2)(n/3) < 1/80.
A little common sense tells us some things about powers of ½.

(1/2)2 = ¼
(1/2)3 = 1/8
(1/2)5 = 1/32
(1/2)6 = 1/64
(1/2)7 = 1/128

Since (1/2)7 < 1/80 < (1/2)6, then n/3 must be somewhere between 6 and 7, giving n somewhere between 18 and 21 hours.

We can narrow this down more with a little work. 

We are interested in solving (1/2)n/3 = 1/80.  Logarithms can help here. 

Whenever the variable we want to find is in the exponent, logs can help release it from that position so that we can find it. 

Here is how:

log (1/2)n/3 = log 1/80           (because if two quantities are equal, so are their
                                                  logs).
(n/3)log (1/2) = log 1/80       (this is a well-known property of logarithms)

n/3 = (log 1/80) / log(1/2)    so,

n = 3 (log 1/80) / log (1/2);  that is, n is approximately 18.97 hours, or about 19
                                            hours.  Our hunch was right!