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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?

Sagot :

Answer: 6.93 years

Step-by-step explanation:

Given

Rate of interest is [tex]r=10\%[/tex]

Future value is given by [tex]s=1000e^{0.1t}[/tex]

For the investment to double itself, i.e. [tex]s=2000[/tex]

[tex]\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}[/tex]

It takes around 6.93 years to double the investment.

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