Question

Compound interest of $1000 is invested at 10% compounded continuously, the future value s at any time t in years is given by s=1000 e^0.1t how log before the investment doubles?

Answer 1

Answer: 6.93 years

Explanation:

Given

Rate of interest is
`r=10\%`

Future value is given by
`s=1000e^(0.1t)`

For the investment to double itself, i.e.
`s=2000`

`\Rightarrow 2000=1000e^(0.1t)\\\Rightarrow 2=e^(0.1t)\\\\\text{Taking log both sides}\\\\\Rightarrow \ln 2=0.1t\\\\\Rightarrow t=(\ln 2)/(0.1)\\\\\Rightarrow t=6.93\ \text{years}`

It takes around 6.93 years to double the investment.