A vehicle purchased for $20700 depreciates at a constant rate of 15% each year. The value after $t$ years is given by the equation $A = P(0.85)^t$, where $P$ is the original value of the vehicle. We want to determine the number of years, $t$, it takes for the vehicle to depreciate to $11000. We are given that the initial value $P = 20700$ and the final value $A = 11000$. We need to find the value of $t$.

1. Problem Description

A vehicle purchased for $20700 depreciates at a constant rate of 15$t$years is given by the equation$A = P(0.85)^t$, where$P$is the original value of the vehicle. We want to determine the number of years,$t$, it takes for the vehicle to depreciate to$

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0. We are given that the initial value $P = 20700$ and the final value $A = 11000$. We need to find the value of $t$.

2. Solution Steps

We are given the equation $A = P(0.85)^t$. We know $A = 11000$ and $P = 20700$.

Substitute these values into the equation:

$11000 = 20700(0.85)^t$

Divide both sides by $20700$:

$\frac{11000}{20700} = (0.85)^t$

$\frac{110}{207} = (0.85)^t$

Take the natural logarithm of both sides:

$\ln(\frac{110}{207}) = \ln((0.85)^t)$

Use the power rule of logarithms: $\ln(a^b) = b\ln(a)$.

$\ln(\frac{110}{207}) = t \ln(0.85)$

Solve for $t$:

$t = \frac{\ln(\frac{110}{207})}{\ln(0.85)}$

$t = \frac{\ln(110/207)}{\ln(0.85)} \approx \frac{-0.6377}{-0.1625} \approx 3.924$

Round to the nearest tenth:

$t \approx 3.9$

3. Final Answer

3.9