sydneykenji12
Answered

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Camden is going to invest $600 and leave it in an account for 12 years. Assuming the
interest is compounded continuously, what interest rate, to the nearest hundredth of
a percent, would be required in order for Camden to end up with $920?

Sagot :

[tex]~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$920\\ P=\textit{original amount deposited}\dotfill & \$600\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &12 \end{cases}[/tex]

[tex]920=600e^{\frac{r}{100}\cdot 12}\implies \cfrac{920}{600}=e^{\frac{3r}{25}}\implies \cfrac{23}{15}=e^{\frac{3r}{25}}\implies \log_e\left( \cfrac{23}{15} \right)=\log_e\left( e^{\frac{3r}{25}} \right) \\\\\\ \ln\left( \cfrac{23}{15} \right)=\cfrac{3r}{25}\implies 25\cdot \ln\left( \cfrac{23}{15} \right)=3r\implies \cfrac{25\cdot \ln\left( \frac{23}{15} \right)}{3}=r \\\\\\ 3.5620\approx r\implies \stackrel{\%}{3.56}\approx r[/tex]

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