Question 1080312: A computer that, when purchased 5 years ago cost $5,000 now has a value of $1,100. Find the value of the computer after 8 years by using the exponential model Vof(t) = V0ekt, in which Vof(t) is the value of the computer at any time t, V0 is the initial cost, and t is the time in years. Round your answer to the nearest hundredth.
Any help is greatly appreciated!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! lets change the variables so they're easier to type and distinguish from each other.
the formula you are looking for is f = p * e^(kt)
f is the future value
p is the present value
e is the scientific constant of 2.718281828.....
k is the rate per time period which is expressed in years.
t is the number of time periods which is expressed in years.
you are given that, 5 years ago, the computer cost $5,000.
you are given that it now has a value of $1,1000.
you want to find the value of the computer after 8 years, which is presumably 3 years from now.
first you want to solve for k.
when f = 1100 and p = 5000 and t = 5, the formula becomes:
1100 = 5000 * e^(k*5) which can also be written as:
1100 = 5000 * e^(5k)
divide both sides of this equaiton by 5000 to get:
1100 / 5000 = e^(5k)
take the natural log of both sides of the equaton to get:
ln(1100/5000) = ln(e^(5k))
since ln(e^x) = x*ln(e) and since ln(e) is equal to 1, your formula becomes:
ln(1100/5000) = 5k
divide both sides of this equation by 5 and solve for k to get:
k = ln(1100/5000) / 5 = -.3028255465
confirm that's correct by replacing k with that value in the original equation to get:
1100 = 5000 * e^(-.3028255465 * 5) which results in 1100 = 1100 which is true, confirming the solution is correct.
now that you know the value of k, you can solve for 8 years instead of 5.
your formula of f = p * e^(kt) becomes f = 5000 * e^(-.3028255465 * 8) which becomes f = 443.4516689
since 8 is 3 more than 5, you could also have used the same formula with a value of 3 for t and a value of 1100 for p to get:
f = p * e^(kt) becomes f = 1100 * e^(-.3028255465 *3) which becomes f = 443.4516689 which is the same answer you got when you did it starting with 5000 eight years ago instead of 1100 three years ago.
k becomes your constant of variation which never changes.
k is also called r in some equation because it is the rate of change per time period.
in your terminology, i think the formula would look like this:
v(t) = v(0) * e^(kt)
v(t) equates to f in my formula
v(0) equates to p in my formula
what you show as ekt is really e^(kt) which means e raised to the power of k * t.
when t = 5, the formula becomes:
v(5) = v(0) * e^(5k)
once you know the value of k, you can then solve the problem for 5, 8, and 3 years.
with 5 years, the formula remains the same at v(5) = v(0) * e^(5k)
with 8 years from 5 years ago, the formula becomes v(8) = v(0) * e^(8k)
with 3 years from now, the formula becomes v(8) = v(5) * e^(3k)
the solutions were:
v(0) = 5000
v(5) = 1100
v(8) = 443.4516689
i think i got it right.
check it out and come back with questions if you still have any.
email dtheophilis@gmail.com
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