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Find the required annual interest rate, to the
nearest tenth of a percent, for $1100 to grow to
$1400 if interest is compounded monthly for 7
years.

Sagot :

[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$1400\\ P=\textit{original amount deposited}\dotfill &\$1100\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &7 \end{cases}[/tex]

[tex]1400=1100\left( 1+\frac{r}{12} \right)^{12\cdot 7}\implies \cfrac{1400}{1100}=\left( 1+\frac{r}{12} \right)^{84}\implies \cfrac{14}{11}=\left( 1+\frac{r}{12} \right)^{84} \\\\\\ \sqrt[84]{\cfrac{14}{11}}=1+\cfrac{r}{12}\implies \sqrt[84]{\cfrac{14}{11}}-1=\cfrac{r}{12}\implies 12\left[ \sqrt[84]{\cfrac{14}{11}}-1 \right]=r \\\\\\ 0.0345\approx r\stackrel{\textit{converting to \%}}{0.0345\cdot 100}\implies 3.45\approx r\implies \stackrel{\textit{rounded up}}{3.4\approx r}[/tex]