The dollar value \( v(t) \) of a certain car model that is \( t \) years old is given by the following exponential function. \[ v(t)=27,500(0.80)^{t} \] Find the value of the car after 7 years and after 13 years. Round your answers to the nearest dollar as necessary.

Calculate or simplify the expression \( 27500*(0.8)^7 \). Calculate the value by following steps: - step0: Calculate: \(27500\times 0.8^{7}\) - step1: Convert the expressions: \(27500\left(\frac{4}{5}\right)^{7}\) - step2: Simplify: \(27500\times \frac{4^{7}}{5^{7}}\) - step3: Evaluate the power: \(27500\times \frac{16384}{5^{7}}\) - step4: Rewrite the expression: \(625\times 44\times \frac{16384}{5^{7}}\) - step5: Rewrite the expression: \(5^{4}\times 44\times \frac{16384}{5^{7}}\) - step6: Reduce the numbers: \(44\times \frac{16384}{5^{3}}\) - step7: Multiply: \(\frac{44\times 16384}{5^{3}}\) - step8: Multiply: \(\frac{720896}{5^{3}}\) - step9: Simplify: \(\frac{720896}{125}\) Calculate or simplify the expression \( 27500*(0.8)^13 \). Calculate the value by following steps: - step0: Calculate: \(27500\times 0.8^{13}\) - step1: Convert the expressions: \(27500\left(\frac{4}{5}\right)^{13}\) - step2: Simplify: \(27500\times \frac{4^{13}}{5^{13}}\) - step3: Rewrite the expression: \(625\times 44\times \frac{4^{13}}{5^{13}}\) - step4: Rewrite the expression: \(5^{4}\times 44\times \frac{4^{13}}{5^{13}}\) - step5: Reduce the numbers: \(44\times \frac{4^{13}}{5^{9}}\) - step6: Multiply: \(\frac{44\times 4^{13}}{5^{9}}\) The value of the car after 7 years is $5767.168 and after 13 years is $1511.828.