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A car is purchased for $31,000. Each year it loses 25% of its value. After how many years will the car be worth $6400 or less?

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A car is purchased for 31,000.Eachyearitloses2531,000. Each year it loses 25% of its value. After how many years will the car be worth 6400 or less?

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Solution 1

Sure, here is the step by step solution:

Step 1: Understand the problem. The car's value decreases by 25% each year. We need to find out when it will be worth $6400 or less.

Step 2: Set up the equation. The formula for depreciation is:

Value = Initial Value * (1 - Depreciation Rate) ^ Number of Years

In this case, the depreciation rate is 25% or 0.25, the initial value is 31,000,andwewanttofindoutwhenthevaluewillbe31,000, and we want to find out when the value will be 6400 or less.

So, we set up the equation as follows:

6400 >= 31000 * (1 - 0.25) ^ Number of Years

Step 3: Solve the equation. This is a logarithmic equation, so we'll need to use logarithms to solve it.

First, divide both sides by 31000 to simplify:

0.20645 >= (0.75) ^ Number of Years

Then, take the natural logarithm (ln) of both sides:

ln(0.20645) >= Number of Years * ln(0.75)

Finally, divide both sides by ln(0.75) to solve for the number of years:

Number of Years >= ln(0.20645) / ln(0.75)

Step 4: Calculate the result. Using a calculator, we find that:

Number of Years >= 7.99

Since we can't have a fraction of a year, we round up to the nearest whole number.

So, the car will be worth $6400 or less after 8 years.

This problem has been solved

Solution 2

Sure, here is the step by step solution:

  1. We start with the initial value of the car which is $31,000.

  2. Each year, the car loses 25% of its value. This means that each year, the car's value is 75% of its value the previous year.

  3. We need to find out how many years it will take for the car's value to be $6400 or less.

  4. We can set up the equation 31000*(0.75)^n = 6400, where n is the number of years.

  5. To solve for n, we can take the natural logarithm (ln) of both sides of the equation to get: ln(31000*(0.75)^n) = ln(6400).

  6. Using the properties of logarithms, we can simplify this to: n*ln(0.75) = ln(6400) - ln(31000).

  7. Solving for n gives: n = (ln(6400) - ln(31000)) / ln(0.75).

  8. Using a calculator, we find that n ≈ 8.67.

  9. Since we can't have a fraction of a year, we round up to the nearest whole number.

So, it will take approximately 9 years for the car to be worth $6400 or less.

This problem has been solved

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