Question

Bella invested $680 in an account paying an interest rate of 2 3/8% compounded

annually. Tanisha invested $680 in an account paying an interest rate of 2 7/8%
compounded quarterly. To the nearest hundredth of a year, how much longer would
it take for Bella's money to triple than for Tanisha's money to triple?

Answer 1

well, first off let's check on Bella, 2 and 3/8, that'd be a rate of 2.375%, well 680 tripled is simply 2040, so let's check

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$2040\\ P=\textit{original amount deposited}\dotfill &\$680\\ r=rate\to 2.375\%\to (2.375)/(100)\dotfill &0.02375\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years \end{cases}`

`2040=680\left(1+(0.02375)/(1)\right)^(1\cdot t) \implies \cfrac{2040}{680}=1.02375^t\implies 3=1.02375^t \\\\\\ \log(3)=\log(1.02375^t)\implies \log(3)=t\log(1.02375) \\\\\\ \cfrac{\log(3)}{\log(1.02375)}=t\implies {\Large \begin{array}{llll} 46.805\approx t \end{array}}`

now let's check on Tanisha, where 2 and 7/8 will be a rate of 2.875%, same tripled amount of 2040

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$2040\\ P=\textit{original amount deposited}\dotfill &\$680\\ r=rate\to 2.875\%\to (2.875)/(100)\dotfill &0.02875\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years \end{cases}`

`2040=680\left(1+(0.02875)/(4)\right)^(4\cdot t) \implies \cfrac{2040}{680}=1.0071875^(4t)\implies 3=1.0071875^(4t) \\\\\\ \log(3)=\log(1.0071875^(4t))\implies \log(3)=t\log(1.0071875^4) \\\\\\ \cfrac{\log(3)}{\log(1.0071875^4)}=t\implies {\Large \begin{array}{llll} 38.350\approx t \end{array}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{Bella's}{46.805}~~ - ~~\stackrel{Tanisha's}{38.350} ~~ \approx ~~ 8.455~~ \approx ~~\text{\LARGE 8.46} ~~ \textit{about 8 years and 168 days}`