Question

An amount of money is deposited in an account that pays 8% annual interest that is compounded 4 times per year.

How long will it take for the amount to triple? (round to 3 decimal places)
How long will is take for the amount to triple if the interest is compounded continuously? (round to three decimal places)

Answer 1

let's take it from the basic one dolla!! So how long for one dolla to turn into 3?

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 3\\ P=\textit{original amount deposited}\dotfill &\$1\\ r=rate\to 8\%\to (8)/(100)\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year} \end{array}\dotfill &4\\ t=years \end{cases}`

`3 = 1\left(1+(0.08)/(4)\right)^(4\cdot t) \implies 3=1.02^(4t)\implies \log(3)=\log(1.02^(4t)) \\\\\\ \log(3)=t\log(1.02^(4))\implies \cfrac{\log(3)}{\log(1.02^(4))}=t\implies \stackrel{years}{13.870\approx t} \\\\[-0.35em] ~\dotfill`

`~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 3\\ P=\textit{original amount deposited}\dotfill & \$1\\ r=rate\to 8\%\to (8)/(100)\dotfill &0.08\\ t=years \end{cases} \\\\\\ 3 = 1e^(0.08\cdot t) \implies \log_e(3)=\log_e(e^(0.08t))\implies \log_e(3)=0.08t \\\\\\ \ln(3)=0.08t\implies \cfrac{\ln(3)}{0.08}=t\implies \stackrel{years}{13.733\approx t}`