Jonas purchased a new car for $25,000. Each year the value of the car depreciates by 20% of its value the previous year. In how many years will the car be worth $5000?
25000(.8)^n = 5000
.8^n = .2
take log of both sides , and using log rules
n log.8 = log.2
n = log .2/log .8 = appr 7.2 years
This is actually a geometric series problem...which has the formula
An=(A1)(r)^(n-1)
so substitute the given
r=1-1/5 = 4/5
5000 = (25000)(4/5)^(n-1)
5000/25000 = (4/5)^(n-1)
1/5 = (4/5)^(n-1)
take ln on both sides
ln(1/5) = ln[(4/5)^(n-1)]
ln(1/5) = (n-1)ln(4/5)
n = [ln(1/5)/ln(4/5)]+1
n = 8.21 years
To find out in how many years the car will be worth $5000, we can set up a mathematical equation to represent the depreciation of the car's value over time.
Let's assume that the car's value after n years is V. We know that the car depreciates by 20% of its value each year, so the new value each year can be calculated as 0.8 times the previous value.
Therefore, we can write the equation as follows:
V = 25000 * (0.8)^n
We want to find the value of n when V is equal to $5000. So we can set up the equation:
5000 = 25000 * (0.8)^n
To solve this equation for n, we can use logarithms as follows:
Take the logarithm on both sides of the equation:
log(5000) = log(25000 * (0.8)^n)
Using the property of logarithms, we can simplify the equation:
log(5000) = log(25000) + n * log(0.8)
Now, substitute the values into the equation and solve for n:
log(5000) = log(25000) + n * log(0.8)
log(5000) = log(25000) + n * (-0.09691) (log(0.8) ≈ -0.09691)
log(5000) = 4.39794 + (-0.09691n) (log(25000) ≈ 4.39794)
Rearrange the equation:
-0.09691n = log(5000) - 4.39794
-0.09691n = 3.69897 - 4.39794
-0.09691n = -0.69897
Divide both sides of the equation by -0.09691:
n = -0.69897 / -0.09691
Calculating the result:
n ≈ 7.2006
Since the number of years cannot be in decimal, we round up to the nearest whole number. Therefore, the car will be worth $5000 after approximately 8 years.