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A new car is purchased for 17300 dollars. The value of the car depreciates at
6.5% per year. To the nearest year, how long will it be until the value of the car is 10600 dollars?
Answer:

A new car is purchased for $17300$ dollars. The value of the car depreciates at $6.5 \%$ per year. To the nearest year, how long will it be until the value of the car is $10600$ dollars? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A new car is purchased for

17300

dollars. The value of the car depreciates at

6.5 \%

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

10600

dollars?

\newline

Answer:

Identify Values: Identify the initial value, the rate of depreciation, and the final value. $\newline$ The initial value $P_0$ of the car is $\$17,300$ , the rate of depreciation $r$ is $6.5\%$ per year, and the final value $P$ we want to find the time for is $\$10,600$ .
Set Up Formula: Set up the formula for exponential decay. $\newline$ The formula for exponential decay is $P = P_0 \times (1 - r)^t$ , where $P$ is the final amount, $P_0$ is the initial amount, $r$ is the rate of depreciation, and $t$ is the time in years.
Substitute Values: Substitute the known values into the formula. $\newline$ $\$10,600 = \$17,300 \times (1 - 0.065)^t$
Solve for t: Solve for t. $\newline$ First, divide both sides by $\$17,300$ to isolate the exponential part of the equation. $\newline$ $\$10,600 / \$17,300 = (1 - 0.065)^t$ $\newline$ $0.6127177570093458 \approx (0.935)^t$
Take Natural Logarithm: Take the natural logarithm of both sides to solve for $t$ . $\ln(0.6127177570093458) = t \times \ln(0.935)$
Calculate t Value: Calculate the value of $t$ . $t = \frac{\ln(0.6127177570093458)}{\ln(0.935)}$ $t \approx 9.574$
Round to Nearest Year: Round the value of $t$ to the nearest year. $\newline$ Since we need to find the number of years to the nearest year, we round $9.574$ to $10$ years.

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