1. 5.4 Applications of Exponential Functions
Compound Interest
Annually
n times per year
Continuously
Decay
½ Life
Growth
Population
Doubling Time

2. Example 1: Different Compounding Periods
Determine the amount that a $5000 investment over ten years at an annual interest rate of 4.8% is worth for each compounding period.
A. annually A = 5000(1+.048)10=$
B. quarterly A = 5000(1+.048/4)10(4)=$
C. monthly A = 5000(1+.048/12)10(12)=$
D. continuously A = 5000(e)10(.048)=$

3. Example 2: Solving for the Time Period
If $7000 is invested at 5% annual interest, compounded monthly, when will the investment be worth $8500?
8500=7000(1+.05/12)12t
t = years
This is equivalent to a 5.116% interest rate compounded annually
( ) 12 = =5.116%

4. Example 3
If f(x) has a continuous growth rate of .17, what is the yearly percentage growth rate?
If f(x) has a monthly percentage growth rate of 3.1%, what is its continuous growth rate?
𝐴=𝑃 𝑒 𝑟𝑡 𝑣𝑠 𝐴=𝑃(1+𝑟 ) 𝑡
𝑒 = =
𝑟=18.530%
𝐴=𝑃(1+ 𝑟 12 ) 12𝑡 𝑣𝑠 𝐴=𝑃 𝑒 𝑟𝑡
( ) = 𝑒 𝑟
= 𝑒 𝑟
𝑟=𝑙𝑛 = =36.635%

5. Example 4: Population Growth
The population of Tokyo, Japan, in the year 2000 was about 26.4 million and is projected to increase at a rate of approximately 0.19% per year. Write the function that gives the population of Tokyo in year x, where x=0 corresponds to 2000.
f(x)=26.4(1.0019)x

6. Example 5: Population Growth
A newly formed lake is stocked with 900 fish. After 6 months, biologists estimate there are 1710 fish in the lake. Assuming the fish population grows exponentially, how many fish will there be after 24 months?

7. Example 5: Solution
Using the formula f(x)=abx, we have f(x)=900bx, which leaves us to determine b. Use the other given data about the population growth to determine b.
In 6 months, there were 1710 fish:
f(6)=  = 900b6  b=1.91/6
f(x)=900(1.91/6)x
f(24)= 900(1.91/6)24 = 11,728.89
There will be about 11,729 fish in the lake after 24 months.

8. Example 6: Chlorine Evaporation
Each day, 15% of the chlorine in a swimming pool evaporates. After how many days will 60% of the chlorine have evaporated?

9. Example 6: Solution
Since 15% of the chlorine evaporates each day, there is 85% remaining. This is the rate, or b value.
f(x)=(1-0.15)x=0.85x
We want to know when 60% has evaporated; or, when there is 40% left.
*Let a = 1
f(x)=0.85x
0.40=0.85x
X= 5.368

10. Example 7: ½ life
Suppose that the half-life of the fictional radioactive substance unobtainium is 29 years. A sample of rock is found with 80 grams of unobtainium. What is the percentage by which the amount of unobtainium in the rock decreases each year?
𝐴=𝑃( 1 2 ) 𝑡 29
1−.97638=.02362
The % of decrease is 2.362%
𝐴=𝑃 ( 1 2 ) ) 𝑡
𝑏=(.5 ) ≈.97638
THIS IS THE % THAT REMAINS!

11. Example 8: Doubling Time
Suppose a population of bacteria is growing in a petri dish in such a way that the number of bacteria doubles every 3 hours. If there are initially 1000 bacteria present, how long will it take for there to be 1 million bacteria?
𝐴=𝑃(2 ) 𝑡 3
1,000,000=1000(2 ) 𝑡 3
1,000=(2 ) 𝑡 3
𝑡= 3 𝑙𝑛1,000 𝑙𝑛2 ≈ ℎ𝑜𝑢𝑟𝑠