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A new car is purchased for 17600 dollars. The value of the car depreciates at
7.75% per year. To the nearest tenth of a year, how long will it be until the value of the car is 6ooo dollars?
Answer:

A new car is purchased for $17600$ dollars. The value of the car depreciates at $7.75 \%$ per year. To the nearest tenth of a year, how long will it be until the value of the car is $6$ ooo dollars? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A new car is purchased for

17600

dollars. The value of the car depreciates at

7.75 \%

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

6

ooo dollars?

\newline

Answer:

Determine formula for depreciation: Determine the formula for depreciation. $\newline$ The car depreciates at a constant percentage rate per year, which is an exponential decay problem. $\newline$ The formula for exponential decay is $P(t) = P_0 \cdot e^{(-kt)}$ , where $P(t)$ is the value at time $t$ , $P_0$ is the initial value, $k$ is the decay constant, and $e$ is the base of the natural logarithm.
Calculate decay constant $k$ : Calculate the decay constant $k$ . The decay constant $k$ can be found using the annual depreciation rate of $7.75\%$ . $k = 0.0775$ (since $7.75\% = \frac{7.75}{100} = 0.0775$ )
Set up equation with given values: Set up the equation with the given values. $P_0 = \$17,600$ , $P(t) = \$6,000$ , and $k = 0.0775$ . $\$6,000 = \$17,600 \times e^{(-0.0775t)}$
Solve for t: Solve for t. $\newline$ Divide both sides by $\$17,600$ to isolate the exponential term. $\newline$ $\$6,000 / \$17,600 = e^{-0.0775t}$ $\newline$ $0.3409 \approx e^{-0.0775t}$
Take natural logarithm to solve for t: Take the natural logarithm of both sides to solve for t. $\newline$ $\ln(0.3409) = \ln(e^{-0.0775t})$ $\newline$ $\ln(0.3409) = -0.0775t$
Divide by $-0.0775$ to find $t$ : Divide by $-0.0775$ to find $t$ .
$t = \frac{\ln(0.3409)}{-0.0775}$
$t \approx \frac{\ln(0.3409)}{-0.0775}$
$t \approx 14.206$ (to the nearest tenth)

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