Question 1080312
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