Question

A person invests 7000 dollars in a bank. The bank pays 5. 5% interest compounded monthly. To the , how long must the person leave the money in the bank until it reaches 13200 dollars?

A=P(1+r/n)^nt

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Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 13200\\ P=\textit{original amount deposited}\dotfill &\$7000\\ r=rate\to 5.5\%\to (5.5)/(100)\dotfill &0.055\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}`

`13200 = 7000\left(1+(0.055)/(12)\right)^(12\cdot t) \implies \cfrac{13200}{7000}=\left(1+(0.055)/(12)\right)^(12t)`

`\cfrac{66}{35}=\left((12+0.055)/(12)\right)^(12t)\implies \cfrac{66}{35}=\left((12.055)/(12)\right)^(12t)\implies \log\left( \cfrac{66}{35} \right)=\log\left[ \left((12.055)/(12)\right)^(12t) \right] \\\\\\ \log\left( \cfrac{66}{35} \right)=t\log\left[ \left((12.055)/(12)\right)^(12) \right]\implies \cfrac{\log\left( (66)/(35) \right)}{\log\left[ \left((12.055)/(12)\right)^(12) \right]}=t\implies \boxed{11.56\approx t}`