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

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=$7990.66 B. quarterly A = 5000(1+.048/4)10(4)=$8057.32 C. monthly A = 5000(1+.048/12)10(12)=$8072.64 D. continuously A = 5000(e)10(.048)=$8080.37

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 = 3.891 years This is equivalent to a 5.116% interest rate compounded annually (1+ .05 12 ) 12 =1.051162=5.116%

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 .17 = 1.18530 = 1+.18530 𝑟=18.530% 𝐴=𝑃(1+ 𝑟 12 ) 12𝑡 𝑣𝑠 𝐴=𝑃 𝑒 𝑟𝑡 (1+.031 ) 12 = 𝑒 𝑟 1.44246 = 𝑒 𝑟 𝑟=𝑙𝑛1.44246= .36635=36.635%

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

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?

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)=1710  1710= 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.

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?

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

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 ) 1 29 ) 𝑡 𝑏=(.5 ) 1 29 ≈.97638 THIS IS THE % THAT REMAINS!

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 ≈29.897 ℎ𝑜𝑢𝑟𝑠