Question 1145684: This is a Value Depreciation Model
A company purchases a complete industrial networking system for $32,145. The value of the system depreciates at a rate of 7.5% per year. How much will it be worth in 5.5 years, and how long will it take until the system is worth $10,000?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the purchase price is 32,145.
it depreciates at the rate of 7.5% per year.
the formula to use is f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the interest rate per time period
n is then number of time periods.
the time periods are years.
in your problem:
f = what you wand to find
p = 32,145
r = -7.5% = -.075
n = 5.5
formula becomes f = 32,145 * (1 - .075) ^ 5.5
solve for f to get f = 20,935.96387
that's the value after 5.5 years.
to determine how long until the value drops to 10,000, the same formula is used.
f = 10,000
p = 32,145
r = -7.5% per year = -.075
n = what you want to find.
the formula becomes 10,000 = 32,145 * (1 - .075) ^ n
divide both sides of this equation by 32,145 to get:
10,000 / 32,145 = (1 - .075) ^ n
take the log of both sides of this equation to get:
log(10,000 / 32,145) = log((1 - .075) ^ n)
since log((1 - .075) ^ n) is equal to n * log(1 - .075), you get:
log(10,000 / 32,145) = n * log(1 - .075)
divide both sides of this equation by log(1 - .075) to get:
log(10,000 / 32,145) / log(1 - .075) = n
solve for n to get n = 14.97753639.
confirm by replacing n in the original equation with that to get:
10,000 = 32,145 * (1 - .075) ^ 14.97753639.
evaluate that equation to get:
10,000 = 10,000
this confirms the value of n is correct.
your solution is that it will take 14.97753639 years for the original value to become 10,000 dollars.
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