Question

As a promotion, a clothing store draws the name of one of its customers each week. The prize is a coupon for the store. If the winner is not present at the drawing, he or she cannot claim the prize, and the amount of the coupon increases for the following week's drawing. The function f(x) = 20 (1.2)^x gives the amount of the coupon in dollars after x weeks of the prize going unclaimed. {{ x is the exponent on (1.2). }}
(A) What is the percent increase each week?

(B) What is the original amount of the coupon? (the initial value)

(C) What is the amount of the coupon after 2 weeks of the prize going unclaimed?

(D) After how many weeks of the prize going unclaimed will the amount of the coupon be greater than $100?

Answer 1

The given formula is f(x) = 20(1.2)^x

The formula is the starting amount multiplied by 1 + the percentage raised to the number of weeks.

A) the percent increase is 20% ( 1.2 in the formula is 1 +20% as a decimal)

B) the original amount is $20

C) for 2 weeks, replace x with 2 and solve:

20(1.2)^2

20(1.44) = $28.80

After 2 weeks the coupon is $28.80

D)  To solve for the number of weeks (x) set the equation equal to $100:

100 = 20(1.2)^x

Divide both sides by 20:

5 = 1.2^x

Take the natural logarithm of both sides:

ln(5) = ln(1.2^x)

Use the logarithm rule to remove the exponent:

ln(5) = x ln(1.2)

Divide both sides by ln(1.2)

x = ln(5) / ln(1.2)

Divide:

X = 8.83

At 8.83 weeks the coupon would be $100, so after 9 weeks the coupon would be greater than $100

The answer is 9 weeks.