A vehicle purchased for $ 20700 depreciates at a constant rate of 18 % each year. Its value after t years can be modeled by the equation A

To solve for the number of years, $t$, it takes for the vehicle's value to depreciate to $8000, we can set up the equation based on the given model:

$A = P \cdot (0.82)^t$

where:

• $A = 8000$ (the target depreciated value),
• $P = 20700$ (the original value of the vehicle),
• and $0.82$ is the annual depreciation factor.

Substitute the known values into the equation:

$8000 = 20700 \cdot (0.82)^t$

Now, solve for $t$:

1. Divide both sides by 20700 to isolate $(0.82)^t$:

   $\frac{8000}{20700} = (0.82)^t$

2. Calculate the left side:

   $0.3865 \approx (0.82)^t$

3. Take the natural logarithm of both sides to bring down the exponent:

   $\ln(0.3865) = t \cdot \ln(0.82)$

4. Solve for $t$ by dividing both sides by $\ln(0.82)$:

   $t = \frac{\ln(0.3865)}{\ln(0.82)}$

5. Compute the value of $t$:

   $t \approx \frac{-0.9512}{-0.1987} \approx 4.8$

So, it takes approximately 4.8 years for the vehicle to depreciate to $8000, rounded to the nearest tenth.

Would you like further details, or do you have any questions about this calculation?

Here are some related questions that might deepen your understanding:

1. What would the value of the vehicle be after 3 years?
2. How does changing the depreciation rate affect the time it takes to reach a specific value?
3. If the depreciation rate were 20% instead of 18%, how many years would it take to reach $8000?
4. What would the equation look like if the vehicle’s value halved every 3 years?
5. How could you calculate the vehicle’s depreciation value over non-integer years, such as 4.5 years?

Tip: Depreciation problems often use exponential decay functions, so understanding logarithmic transformations can be very helpful in solving for time