Question

An initial investment of $1000 is appreciated for 8 years in an account that earns 9% interest, compounded annually. Find the amount of money in the account at the end of the period.
Let f(x) = 3x. Find f( 2).
Solve the equation 1296 x = 6

Answer 1

These are three different questions. I will solve eat one in a separated post.

Answer 2

The amount of money by the end of the 8th year is \(1000(1+0.09)^8=$1993\).

Answer 3

For the second question, just plug \(x=2\) to find f(2). That's \(f(2)=3(2)=6\).

Answer 4

In the last problem, divide both sides by \(1296\) to get x. \(x=\frac{6}{1296}=\frac{1}{216}\).

Answer 5

3 separate questions?

1: annually compounding interest formula FV=P(1+i/r)^(rt)
FV = Future Value
P = Present Value
i = interest rate
r = compounding frequency (annually, monthly, so on)
t = time (interest periods how many periods pass

so, in your case:
FV=1000(1+(.09))^(8)
FV=1992.56

2. f(2)=6 because you substitute 2 in for x and 2*3 is six

3. x=6/1296

Answer 6

yeah for questions 2 and 3, I thought to answer them so but wasnt sure if they had anything to do with question 1

Answer 7

thanks!