Question

As a promotion, a store draws the name of a customers each week. The prize is a coupon for the store. If winner is not present at drawing, he cannnot claim the prize and the amount of the coupon increases for the next weeks drawing. the function f(x)= 20(1.2)to the x power, gives the amt of the coupon in dollars after x weeks of the prize going unclaimed.
a. what is the amt of the coupon after 2 wks of prize not claimed?
b.after how many weeks of the prize going unclaimed will the amt of the coupon be greater than $100?
c. What is the original amt of the coupon?
d. Find the percent increase each week?

I am having trouble solving d. Please help

Answer 1

A. Since f(x)=20(1.2)^x, substitute x for 2 as for 2 weeks. f(2) = 20(1.2)^2 = 20(1.44) = $28.80.
B. Using the guess and check method on a table, f(x)=20(1.2)^x, f(0)=20, f(1)=24, f(2)=28.8, f(3)=34.56, f(4)=41.472, f(5)=49.7664, f(6)=59.71968, f(7)=71.663616, f(8)=85.9963392, f(9)=103.195607, and 103.195607>100. So, it will take 9 weeks for the prize to be greater than $100 going unclaimed.
C. From f(x)=a(1+r)^x, where a is the initial value and r is the rate of change, and f(x)=20(1.2)^x, $20 would be the original amount of the coupon.
D. From f(x)=a(1+r)^x, where a is the initial value and r is the rate of change, and f(x)=20(1.2)^x, 20% would be the percent increase each week.