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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.

Respuesta :

beritop1089

Answer:

Step-by-step explanation:

Future value with continuous compounding ; FV =[tex]Pe^{rt\\ } \\

900,000 = 15,000e^{38r} \\ \\ \frac{900,000}{15,000}  =e^{38r\\ }\\ \\ \\ 60 =e^{38r}[/tex]

Introduce ln on both sides of the equation

ln 60 = ln[tex]e^{38r}[/tex]

note:  ln[tex]e^{x} = x[/tex]

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.

Answer:

10.8%

Step-by-step explanation:

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