A new car is purchased for 17900 dollars. The value of the car depreciates at \( 13.25 \% \) per year. To the nearest tenth of a year, how long will it be until the value of the car is 2800 dollars?

To find out how long it will take for the value of the car to depreciate to $2800, we can use the formula for exponential decay: \[ V(t) = V_0 \times (1 - r)^t \] where: - \( V(t) \) is the value of the car at time \( t \), - \( V_0 \) is the initial value of the car (17900 dollars), - \( r \) is the depreciation rate (13.25% or 0.1325), - \( t \) is the time in years. We want to find the time \( t \) when the value of the car is $2800. So, we set up the equation: \[ 2800 = 17900 \times (1 - 0.1325)^t \] Now, we can solve for \( t \) using logarithms. Solve the equation by following steps: - step0: Solve for \(t\): \(2800=17900\left(1-0.1325\right)^{t}\) - step1: Simplify: \(2800=17900\left(\frac{347}{400}\right)^{t}\) - step2: Swap the sides: \(17900\left(\frac{347}{400}\right)^{t}=2800\) - step3: Multiply by the reciprocal: \(17900\left(\frac{347}{400}\right)^{t}\times \frac{1}{17900}=2800\times \frac{1}{17900}\) - step4: Multiply: \(\left(\frac{347}{400}\right)^{t}=\frac{28}{179}\) - step5: Take the logarithm of both sides: \(\log_{\frac{347}{400}}{\left(\left(\frac{347}{400}\right)^{t}\right)}=\log_{\frac{347}{400}}{\left(\frac{28}{179}\right)}\) - step6: Evaluate the logarithm: \(t=\log_{\frac{347}{400}}{\left(\frac{28}{179}\right)}\) - step7: Transform the expression: \(t=\log_{\frac{347}{400}}{\left(28\right)}-\log_{\frac{347}{400}}{\left(179\right)}\) The solution to the equation is: \[ t = \log_{\frac{347}{400}}{(28)} - \log_{\frac{347}{400}}{(179)} \] Now, we can calculate the value of \( t \) using logarithms. Calculate the value by following steps: - step0: Calculate: \(\log_{\frac{347}{400}}{\left(28\right)}-\log_{\frac{347}{400}}{\left(179\right)}\) The value of \( t \) is approximately 13.05181 years. Therefore, it will take approximately 13 years for the value of the car to depreciate to $2800.