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a new car is purchased for 18500 dollars. the value of the car depreciates at 8% per year. to the nearest tenth of a year, how long will it be until the value of the car is 9100 dollars?

1 Answer

1 vote

Answer:

8.5 years

Explanation:

You want to know the number of years until 18500 depreciates to 9100 at the rate of 8% per year.

Value

The depreciation rate given as a percentage of current value tells you the depreciation is exponential. The formula will be ...

value = (initial value) × (1 - (depreciation rate))^t

where the rate is "per year" and t is in years.

Application

value = 18500·(1 -0.08)^t

9100 = 18500·0.92^t . . . . fill in the value of interest

9100/18500 = 0.92^t . . . . divide by 18500

log(91/185) = t·log(0.92) . . . . take logarithms

t = log(91/185)/log(0.92) ≈ -0.3081/-0.03621 ≈ 8.509

It will be about 8.5 years until the value is $9100.

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Additional comment

The graph shows the solution to ...

18500·0.92^t -9100 = 0

We find it fairly easy to locate an x-intercept, so we wrote the equation in the forms that makes the x-intercept the solution.

a new car is purchased for 18500 dollars. the value of the car depreciates at 8% per-example-1
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