Respuesta :
[tex]\bf \textit{Logarithm of exponentials}\\\\
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)
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[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\to &\$10000\\ P=\textit{original amount deposited}\to &\$4000\\ r=rate\to 9\%\to \frac{9}{100}\to &0.09\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\to &12\\ t=years\end{cases}[/tex]
[tex]\bf 10000=4000\left(1+\frac{0.09}{12}\right)^{12t}\implies \cfrac{10000}{4000}=\left(1+\frac{0.09}{12}\right)^{12t} \\\\\\ \cfrac{5}{2}=\left(1+\frac{0.09}{12}\right)^{12t}\impliedby \textit{now we take \underline{log} to both sides} \\\\\\ log\left( \frac{5}{2} \right)=log\left[ \left(1.0075\right)^{12t} \right]\implies log\left( \frac{5}{2} \right)=12t\cdot log[1.0075] \\\\\\ \cfrac{log\left( \frac{5}{2} \right)}{12\cdot log(1.0075)}=t[/tex]
recall that a log with no base, implies base 10. But anyway, which is about 10.2 or 10 years 2 months and half.
[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\to &\$10000\\ P=\textit{original amount deposited}\to &\$4000\\ r=rate\to 9\%\to \frac{9}{100}\to &0.09\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\to &12\\ t=years\end{cases}[/tex]
[tex]\bf 10000=4000\left(1+\frac{0.09}{12}\right)^{12t}\implies \cfrac{10000}{4000}=\left(1+\frac{0.09}{12}\right)^{12t} \\\\\\ \cfrac{5}{2}=\left(1+\frac{0.09}{12}\right)^{12t}\impliedby \textit{now we take \underline{log} to both sides} \\\\\\ log\left( \frac{5}{2} \right)=log\left[ \left(1.0075\right)^{12t} \right]\implies log\left( \frac{5}{2} \right)=12t\cdot log[1.0075] \\\\\\ \cfrac{log\left( \frac{5}{2} \right)}{12\cdot log(1.0075)}=t[/tex]
recall that a log with no base, implies base 10. But anyway, which is about 10.2 or 10 years 2 months and half.
