The value \( V \) of a certain automobile that is tyears old can be modeled by \( V(t)=14,111(0.81)^{t} \). According to the model, when will the car be worth each of the following amounts? (a) \( \$ 7000 \) (b) \( \$ 6000 \) (c) \( \$ 4000 \) (a) The car will be worth \( \$ 7000 \) after - y years. (Type an integer or a decimal rounded to the nearest tenth as needed.) (b) The car will be worth \( \$ 6000 \) after \( \square \) years. (Type an integer or a decimal rounded to the nearest tenti as needed.) (c) The car will be worth \( \$ 4000 \) after \( \square \) years. (Type an integer or a decimal rounded to the nearest tentiras needed.)

To find when the car will be worth each of the given amounts, we need to solve the equation \( V(t) = 7000 \), \( V(t) = 6000 \), and \( V(t) = 4000 \) for \( t \) using the given model \( V(t) = 14,111(0.81)^t \). Let's solve each equation step by step: (a) \( V(t) = 7000 \): \[ 14,111(0.81)^t = 7000 \] (b) \( V(t) = 6000 \): \[ 14,111(0.81)^t = 6000 \] (c) \( V(t) = 4000 \): \[ 14,111(0.81)^t = 4000 \] Now, let's solve each equation for \( t \). Solve the equation by following steps: - step0: Solve for \(t\): \(14111\times 0.81^{t}=7000\) - step1: Convert the expressions: \(14111\left(\frac{81}{100}\right)^{t}=7000\) - step2: Multiply by the reciprocal: \(14111\left(\frac{81}{100}\right)^{t}\times \frac{1}{14111}=7000\times \frac{1}{14111}\) - step3: Multiply: \(\left(\frac{81}{100}\right)^{t}=\frac{7000}{14111}\) - step4: Take the logarithm of both sides: \(\log_{\frac{81}{100}}{\left(\left(\frac{81}{100}\right)^{t}\right)}=\log_{\frac{81}{100}}{\left(\frac{7000}{14111}\right)}\) - step5: Evaluate the logarithm: \(t=\log_{\frac{81}{100}}{\left(\frac{7000}{14111}\right)}\) - step6: Simplify: \(t=\frac{\log_{\frac{9}{10}}{\left(7000\right)}-\log_{\frac{9}{10}}{\left(14111\right)}}{2}\) The solution to the equation \( V(t) = 7000 \) is \( t \approx 3.326884 \) years. Now, let's solve the equation \( V(t) = 6000 \) for \( t \). Solve the equation by following steps: - step0: Solve for \(t\): \(14111\times 0.81^{t}=6000\) - step1: Convert the expressions: \(14111\left(\frac{81}{100}\right)^{t}=6000\) - step2: Multiply by the reciprocal: \(14111\left(\frac{81}{100}\right)^{t}\times \frac{1}{14111}=6000\times \frac{1}{14111}\) - step3: Multiply: \(\left(\frac{81}{100}\right)^{t}=\frac{6000}{14111}\) - step4: Take the logarithm of both sides: \(\log_{\frac{81}{100}}{\left(\left(\frac{81}{100}\right)^{t}\right)}=\log_{\frac{81}{100}}{\left(\frac{6000}{14111}\right)}\) - step5: Evaluate the logarithm: \(t=\log_{\frac{81}{100}}{\left(\frac{6000}{14111}\right)}\) - step6: Simplify: \(t=\frac{\log_{\frac{9}{10}}{\left(6000\right)}-\log_{\frac{9}{10}}{\left(14111\right)}}{2}\) The solution to the equation \( V(t) = 6000 \) is \( t \approx 4.058423 \) years. Finally, let's solve the equation \( V(t) = 4000 \) for \( t \). Solve the equation by following steps: - step0: Solve for \(t\): \(14111\times 0.81^{t}=4000\) - step1: Convert the expressions: \(14111\left(\frac{81}{100}\right)^{t}=4000\) - step2: Multiply by the reciprocal: \(14111\left(\frac{81}{100}\right)^{t}\times \frac{1}{14111}=4000\times \frac{1}{14111}\) - step3: Multiply: \(\left(\frac{81}{100}\right)^{t}=\frac{4000}{14111}\) - step4: Take the logarithm of both sides: \(\log_{\frac{81}{100}}{\left(\left(\frac{81}{100}\right)^{t}\right)}=\log_{\frac{81}{100}}{\left(\frac{4000}{14111}\right)}\) - step5: Evaluate the logarithm: \(t=\log_{\frac{81}{100}}{\left(\frac{4000}{14111}\right)}\) - step6: Simplify: \(t=\frac{\log_{\frac{9}{10}}{\left(4000\right)}-\log_{\frac{9}{10}}{\left(14111\right)}}{2}\) The solution to the equation \( V(t) = 4000 \) is \( t \approx 5.982603 \) years. Therefore, the car will be worth: (a) $7000 after approximately 3.326884 years. (b) $6000 after approximately 4.058423 years. (c) $4000 after approximately 5.982603 years.