Respuesta :

so, we know the earned interest is 120, and we know the original deposited amount is 900 bucks, so, the accumulated amount will then be 900 + 120 or 1020 bucks.

[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 &\$1020\\ P=\textit{original amount deposited}\to &\$900\\ r=rate\to 6\%\to \frac{6}{100}\to &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years \end{cases}[/tex]

[tex]\bf 1020=900\left(1+\frac{0.06}{1}\right)^{1\cdot t}\implies \cfrac{1020}{900}=(1.06)^t\implies \cfrac{102}{90}=(1.06)^t \\\\\\ log\left(\frac{102}{90} \right)=log(1.06^t)\implies log\left(\frac{102}{90} \right)=t\cdot log(1.06) \\\\\\ \cfrac{log\left(\frac{102}{90} \right)}{log(1.06)}=t\implies 20.92 \approx t[/tex]

Otras preguntas