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A new car is purchased for 22100 dollars. The value of the car depreciates at
5.25% per year. To the nearest tenth of a year, how long will it be until the value of the car is 12600 dollars?
Answer:

A new car is purchased for $22100$ dollars. The value of the car depreciates at $5.25 \%$ per year. To the nearest tenth of a year, how long will it be until the value of the car is $12600$ dollars? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A new car is purchased for

22100

dollars. The value of the car depreciates at

5.25 \%

per year. To the nearest tenth of a year, how long will it be until the value of the car is

12600

dollars?

\newline

Answer:

Determine depreciation type: Determine the type of depreciation. The car depreciates at a constant percentage each year. This indicates exponential decay.
Identify values: Identify the initial value $(a)$ , the rate of depreciation $(r)$ , and the final value $(P(t))$ . $\newline$ Initial value $(a)$ = $\$22,100$ $\newline$ Rate of depreciation $(r)$ = $5.25\%$ or $0.0525$ (as a decimal) $\newline$ Final value $(P(t))$ = $\$12,600$
Set up formula: Set up the exponential decay formula. $\newline$ The formula for exponential decay is $P(t) = a \cdot (1 - r)^t$ , where $P(t)$ is the value after time $t$ , $a$ is the initial value, and $r$ is the rate of decay.
Substitute values: Substitute the known values into the formula. $\newline$ $\$12,600 = \$22,100 \times (1 - 0.0525)^t$
Solve for t: Solve for t. $\newline$ First, divide both sides by $\$22,100$ to isolate the exponential part of the equation. $\newline$ $\$12,600 / \$22,100 = (1 - 0.0525)^t$ $\newline$ $0.5697 \approx (0.9475)^t$
Take natural logarithm: Take the natural logarithm of both sides to solve for $t$ . $\ln(0.5697) = \ln((0.9475)^t)$ $\ln(0.5697) = t \cdot \ln(0.9475)$
Divide by ln: Divide both sides by $\ln(0.9475)$ to solve for $t$ . $t = \frac{\ln(0.5697)}{\ln(0.9475)}$ $t \approx \frac{\ln(0.5697)}{\ln(0.9475)}$ $t \approx \frac{-0.5621}{-0.0542}$ $t \approx 10.373$
Round to nearest tenth: Round the answer to the nearest tenth of a year. $\newline$ $t \approx 10.4$ years

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