Question

Layla invested $380 in an account paying an interest rate of 5 7/8% compounded quarterly. Brandon invested $380 in an account paying an interest rate of 6 1/8% compounded monthly. To the nearest hundredth of a year, how much longer would it take for Layla's money to triple than for Brandon's money to triple

Answer 1

so hmmm tripling $380 we end up with $1140, now, since 7/8 is 0.875, that means that Layla's rate is 5.875, whilst Brandon's is 6.125.

`~~~~~~ \stackrel{ \textit{\LARGE Layla}}{\textit{Compound Interest Earned Amount}} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 1140\\ P=\textit{original amount deposited}\dotfill &\$380\\ r=rate\to 5.875\%\to (5.875)/(100)\dotfill &0.05875\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years \end{cases}`

`1140 = 380\left(1+(0.05875)/(4)\right)^(4\cdot t) \implies \cfrac{1140}{380}=1.0146875^(4t) \\\\\\ 3=1.0146875^(4t)\implies \log(3)=\log(1.0146875^(4t)) \\\\\\ \log(3)=t\log(1.0146875^(4))\implies \cfrac{\log(3)}{\log(1.0146875^(4))}=t\implies \boxed{18.84\approx t} \\\\[-0.35em] ~\dotfill`

`~~~~~~ \stackrel{ \textit{\LARGE Brandon}}{\textit{Compound Interest Earned Amount}} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 1140\\ P=\textit{original amount deposited}\dotfill &\$380\\ r=rate\to 6.125\%\to (6.125)/(100)\dotfill &0.06125\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}`

`1140 = 380\left(1+(0.06125)/(12)\right)^(12\cdot t) \implies \cfrac{1140}{380}=\left( \cfrac{9649}{9600} \right)^(12t)\implies 3=\left( \cfrac{9649}{9600} \right)^(12t) \\\\\\ \log(3)=\log\left[ \left( \cfrac{9649}{9600} \right)^(12t) \right]\implies \log(3)=t\log\left[ \left( \cfrac{9649}{9600} \right)^(12) \right]`

`\cfrac{\log(3)}{ ~~ \log\left[ \left( (9649)/(9600) \right)^(12) \right] ~~ }=t\implies \boxed{17.98\approx t} \\\\[-0.35em] ~\dotfill\\\\ 18.84~~ - ~~17.98 ~~ \approx ~~ \text{\LARGE 0.86}`