Question

Fred decides to think about retirement and invests at the age of 26. He invests $15,000 and hopes the investment will be

worth $900,000 by the time he turns 64. If the interest compounds continuously, approximately what rate of growth will he
need to achieve his goal? Round to the nearest tenth of a percent.

Answer 1

10.8%

Explanation:

Answer 2

Explanation:

Future value with continuous compounding ; FV =
`Pe^(rt\\ ) \\ </p><p>900,000 = 15,000e^(38r) \\ \\ (900,000)/(15,000)  =e^(38r\\ )\\ \\ \\ 60 =e^(38r)`

Introduce ln on both sides of the equation

ln 60 = ln
`e^(38r)`

note: ln
`e^(x) = x`

Next, solve the equation;

4.09 = 38r

Divide both sides by 38 to solve for r;

4.09/38 = r

r= 0.107746 or 10.7746%

Therefore the rate would be 10.775% to the nearest a tenth of a percent.