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Mademuasel [1]
2 years ago
13

Compound interest of $1000 is invested at 10% compounded continuously, the future value s at any time t in years is given by s=1

000 e^0.1t how log before the investment doubles?
Mathematics
1 answer:
crimeas [40]2 years ago
8 0

Answer: 6.93 years

Step-by-step 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=\dfrac{\ln 2}{0.1}\\\\\Rightarrow t=6.93\ \text{years}

It takes around 6.93 years to double the investment.

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