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The problem involves the depreciation of a car's value. Initially, the car costs $41,000, and it loses 20% of its value each year. We are asked to determine after how many years the car's value will be $13,900 or less.

### Step-by-step solution:

1. **Initial value of the car**: $41,000.
2. **Depreciation rate**: 20% each year, meaning after each year the car retains 80% of its value (since $ 100\% - 20\% = 80\% $).

The car's value after $ t $ years can be modeled by the equation:
\[
V(t) = 41000 \times (0.8)^t
\]
where $ V(t) $ is the value of the car after $ t $ years, and $ t $ is the number of years.

We want to find $ t $ such that $ V(t) \leq 13900 $:
\[
41000 \times (0.8)^t \leq 13900
\]

### Solve for $ t $:
\[
(0.8)^t \leq \frac{13900}{41000}
\]
\[
(0.8)^t \leq 0.33902
\]

Taking the natural logarithm of both sides:
\[
t \ln(0.8) \leq \ln(0.33902)
\]
\[
t \geq \frac{\ln(0.33902)}{\ln(0.8)}
\]

Now, I'll calculate the value of $ t $.The calculated value for $ t $ is approximately 4.85 years. Since the question asks for the smallest possible whole number, the car will be worth $13,900 or less after **5 years**.

Would you like further details or have any other questions?

Here are 5 related questions:
1. What is the value of the car after 3 years?
2. How much value does the car lose in the first year?
3. What percentage of its original value is the car worth after 4 years?
4. How can we calculate the car's depreciation if it loses a different percentage each year?
5. If the car lost 15% instead of 20% annually, how long would it take to reach $13,900?

**Tip:** The exponential decay formula $ V(t) = V_0 \times (1 - r)^t $ is useful for modeling depreciation or any other gradual percentage decrease over time.

A car is purchased for $41,000. Each year it loses 20% of its value. After how many years will the car be worth $13,900 or less?

This problem involves calculating the number of years it will take for a car worth $41,000, losing 20% of its value annually, to be worth $13,900 or less. Learn how to solve using exponential decay and logarithmic calculations. Suitable for Grades 9-11.

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