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A new car is purchased for 24300 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 13000 dollars?
Answer:

A new car is purchased for $24300$ 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 $13000$ dollars? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A new car is purchased for

24300

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

13000

dollars?

\newline

Answer:

Determine type of depreciation: Determine the type of depreciation. The car depreciates at a constant percentage each year. This is exponential decay.
Identify values and formula: Identify the initial value $P_0$ , the rate of depreciation $r$ , and the final value $P$ . $P_0 = \$24,300$ , $r = 5.25\%$ or $0.0525$ in decimal form, $P = \$13,000$ .
Set up exponential decay formula: Set up the exponential decay formula. $\newline$ The formula for exponential decay is $P = P_0 \cdot e^{rt}$ , where $P$ is the final amount, $P_0$ is the initial amount, $r$ is the rate of decay, and $t$ is the time in years.
Substitute values and solve: Substitute the known values into the formula and solve for $t$ . $\$13,000 = \$24,300 \times e^{-0.0525t}$
Divide to isolate expression: Divide both sides by $\$24,300$ to isolate the exponential expression. $\newline$ $(13,000 / 24,300) = e^{(-0.0525t)}$
Calculate left side: Calculate the left side of the equation. $\newline$ $(13,000 / 24,300) \approx 0.5349794238683128$
Take natural logarithm: Take the natural logarithm of both sides to solve for $t$ . $\ln(0.5349794238683128) = \ln(e^{-0.0525t})$
Simplify right side: Use the property of logarithms that $\ln(e^x) = x$ to simplify the right side of the equation. $\ln(0.5349794238683128) = -0.0525t$
Calculate natural logarithm: Calculate the natural logarithm of $0.5349794238683128$ . $\newline$ $\ln(0.5349794238683128) \approx -0.626749673254731$
Divide to solve for $t$ : Divide both sides by $-0.0525$ to solve for $t$ . $t \approx \frac{-0.626749673254731}{-0.0525}$
Calculate value of t: Calculate the value of $t$ . $t \approx 11.9380861558044$
Round to nearest tenth: Round the value of $t$ to the nearest tenth of a year. $\newline$ $t \approx 11.9$ years

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