Question

Carlos invests $4540 at a rate of r% per year compound interest. At the end of 10 years he has earned $1328.54 in interest. Calculate the value of r.​

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \stackrel{ 4540+1328.54 }{\$ 5868.54}\\ P=\textit{original amount deposited}\dotfill &\$4540\\ r=rate\to r\%\to (r)/(100)\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &10 \end{cases}`

`5868.54 = 4540\left(1+((r)/(100))/(1)\right)^(1\cdot 10) \implies \cfrac{5868.54}{4540}=\left( 1+\cfrac{r}{100} \right)^(10) \\\\\\ \cfrac{5868.54}{4540}=\left( \cfrac{100+r}{100} \right)^(10)\implies \sqrt[10]{\cfrac{5868.54}{4540}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[10]{\cfrac{5868.54}{4540}}=100+r\implies 100\sqrt[10]{\cfrac{5868.54}{4540}}-100=r\implies \stackrel{ \% }{2.6}\approx r`