the depreciated value, V, of a car can be calculated using V = C91-r)^n, where C is the original value, r is the rate of depreciation per yeat and n is the age of the car in years. how long would it take for a $24,000 car to be reduced to a value of $2500 if the car is depreciationg at a rate of 15% per year?

Question

the depreciated value, V, of a car can be calculated using V = C91-r)^n, where C is the original value, r is the rate of depreciation per yeat and n is the age of the car in years. how long would it take for a $24,000 car to be reduced to a value of $2500 if the car is depreciationg at a rate of 15% per year?

Answer 1

just plug in your values:

24000(1-0.15)^n = 2500

Now solve for n. First step is

.85^n = 2500/24000

Answer 2

To find out how long it would take for a car to be reduced to a value of $2500, we can rearrange the formula:

V = C(1 - r)^n

Given values:
C = $24,000
V = $2,500
r = 15% or 0.15

Substituting the values into the formula:

$2,500 = $24,000(1 - 0.15)^n

Now, we need to solve for n, the age of the car in years. To do this, we can isolate the exponential term:

(1 - 0.15)^n = $2,500 / $24,000

Simplifying:

0.85^n = 0.10417

To solve for n, we can take the logarithm of both sides. Let's use the natural logarithm (ln):

ln(0.85^n) = ln(0.10417)

Using the logarithmic identity ln(a^b) = b * ln(a):

n * ln(0.85) = ln(0.10417)

Now, we can solve for n:

n = ln(0.10417) / ln(0.85)

Using a calculator, we find:

n ≈ 7.89 years

Therefore, it would take approximately 7.89 years for the $24,000 car to be reduced to a value of $2,500 if it is depreciating at a rate of 15% per year.