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Question 495344: I do not know what this would be under because it is a function word problem.
Assuming an investment of $8,100 doubles in value every 7 years. What function rules models the worth of an investment after 28 years? What about after 35 years?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! f = p * 2^(n/7)
f = future value
p = present value
n = number of years in increments of 7
the formula works like this:
p = 8100
f = what you want to find.
n = 28
formula becomes:
f = 8100 * 2^(28/7) which becomes:
f = 8100 * 2^4 which becomes:
f = 8100 * 16 which becomes:
f = 129600
since 8100 doubles every 7 years, then you get:
time point investment
1 8100
7 16200
14 32400
21 64800
28 129600
the difference between each time point is equal to 7 years.
each time point is equivalent to the value of n in the equation.
time point 0 is when you invest the money
time point 7 is when your money doubles.
time point 14 is when your money doubles again.
time point 21 is when your money doubles again.
time point 28 is when your money doubles again.
you can see that the formula will work for any value of n, except we're restricting the value of n to increments of 7 so the exponent will always be an integer.
we don't need to do that, however.
if we pick any other value of n, we can calculate the future value as well.
for example:
if n = 3, then the formula of:
f = 8100 * 2^(n/7) becomes:
f = 8100 * 2^(3/7) which becomes:
f = 8100 * 1.345900193 which becomes:
f = 10901.79156
in fact, we can solve for the yearly interest rate that causes our money to double every 7 years.
that interest rate is calculated as follows:
f = p * (1+x)^7
we let f = 2 and p = 1 to get:
2 = 1 * (1+x)^7 which becomes:
2 = (1+x)^7
we take the 7th root of both sides of our equation to get:
2^(1/7) = 1 + x
we subtract 1 from both sides of our equation to get:
2^(1/7) - 1 = x
we use our calculator to solve for 2^(1/7) to get:
1.104089514 - 1 = x
we simplify to get:
x = .104089514
that's our annual interest rate that will cause our money to double every 7th year.
to test this out, we go back to our future worth formula of:
f = p * (1+x)^n
we let p = 8100
we let x = .104089514
we let n = 28
we get:
f = 8100 * (1.104089514)^28 which becomes:
f = 8100 * 16 which becomes:
f = 129600
we get the same answer.
either of these formulas will work.
the more traditional formula would be the last one i gave you just now, although the first formula i gave you works fine as well.
notice that when n = 3, the formula becomes:
f = 8100 * (1.104089514)^3 which becomes:
f = 8100 * 1.345900193 which becomes:
f = 10901.79156
that's the same answer we got using the other formula, so either formula will get you the same answer.
the function you are looking for is:
f(n) = 8100 * (1.104089514)^n
n is the number of years.
go out 28, or 35, or any multiple of 7 and you'll see that the money doubles each time.
you could also use:
f(n) = 8100 * 2^(n/7)
either one will get you the same answer.
n is the number of years again.
f(n) represents the future value of the investment.
this is called f(n) rather than f(x) because n is the independent variable.
if we wanted to make it f(x), then the formula would have had to be:
f(x) = 8100 * 2^(x/7)
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