The problem states that a vehicle purchased for $20700 depreciates at a constant rate of 13% each year. The value of the vehicle after $t$ years is modeled by the equation $A = P(0.87)^t$, where $P$ is the original value of the vehicle. We need to determine the number of years it takes for the vehicle to depreciate to $9000.

1. Problem Description

The problem states that a vehicle purchased for $20700 depreciates at a constant rate of 13$t$years is modeled by the equation$A = P(0.87)^t$, where$P$is the original value of the vehicle. We need to determine the number of years it takes for the vehicle to depreciate to$

9

0

0

0.

2. Solution Steps

We are given the equation $A = P(0.87)^t$, where $A$ is the value of the vehicle after $t$ years, and $P$ is the initial price. We are given that the initial price $P = 20700$. We want to find the time $t$ when the value $A = 9000$.

Substitute the given values into the equation:

$9000 = 20700(0.87)^t$

Divide both sides by 20700:

$\frac{9000}{20700} = (0.87)^t$

Simplify the fraction:

$\frac{90}{207} = \frac{10}{23} = (0.87)^t$

To solve for $t$, we can take the natural logarithm (ln) of both sides:

$ln(\frac{10}{23}) = ln((0.87)^t)$

Using the logarithm property $ln(a^b) = b * ln(a)$, we have:

$ln(\frac{10}{23}) = t * ln(0.87)$

Now, divide both sides by $ln(0.87)$ to isolate $t$:

$t = \frac{ln(\frac{10}{23})}{ln(0.87)}$

$t = \frac{ln(10/23)}{ln(0.87)}$

Now, we can use a calculator to find the numerical value of $t$:

$ln(10/23) \approx -0.8285$

$ln(0.87) \approx -0.1393$

$t \approx \frac{-0.8285}{-0.1393} \approx 5.947$

Rounding to the nearest tenth gives $t \approx 5.9$.

3. Final Answer

5. 9