If you make an investment of ​$4,500 and the investment earns a continuous annual interest rate of 4.6​%, in how much many years will it be worth ​$7000​?

Answer 1

There is an equation for continuous compounding

FV=PVe^rt

In this problem,

FV=7000

PV=4500

e is a mathematical constant

r=0.046 (percentage as a decimal)

t is the unknown

Plug everything into this equation and solve

7000=4500 e^(0.046t) Divide both sides by 4500

7000/4500=4500/4500 e^(0.046t) Simplify

14/9 = e^(0.046t) Take the ln of both sides

ln(14/9) = ln(e^(0.046t)) The exponent moves to the front due to the logarithmic power rule

ln (14/9) = (0.046t) ln e Simplify ln e =1

ln (14/9) = (0.046t) divide both sides by 0.046 to isolate t

ln (14/9)/0.046 = (0.046t)/0.046

t= 9.6 years

Answer 2

A 4.6% annual rate means money earns 4.6% yearly.

Since it is "compounded" continuously the amount in t years is 4500*e^{t ln 1.046} and you want this to be 7000

7000=4500 e^{t ln(1.046)}

^{ln(70/45) = t ln(1.046)}

^{t= 9.8 (approximately)}