Question

$10,000, time to double is 12 years, what is the annual rate & amount after 10 years?

Answer 1

well, double of 10000 is 20000, so hmmm happens in 12 years, ok

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$20000\\ P=\textit{original amount deposited}\dotfill &\$10000\\ r=rate\to r\%\to (r)/(100)\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &12 \end{cases}`

`20000=10000\left(1+(~~ (r )/(100 ) ~~)/(1)\right)^(1\cdot 12) \implies \cfrac{20000}{10000}=\left( 1+\cfrac{r}{100} \right)^(12) \\\\\\ 2=\left( \cfrac{100+r}{100} \right)^(12)\implies \sqrt[12]{2}=\cfrac{100+r}{100}\implies 100\sqrt[12]{2}=100+r \\\\\\ 100\sqrt[12]{2}-100=r\implies {\LARGE \begin{array}{llll} \stackrel{\%}{5.95}\approx r \end{array}}`

now, what will it be after 10 years assuming we use 5.95%? mind you that's a rounded figure, but close enough.

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$10000\\ r=rate\to 5.95\%\to (5.95)/(100)\dotfill &0.0595\\ n= \begin{array}{llll} \textit{times it compounds per year} \end{array}\dotfill &1\\ t=years\dotfill &10 \end{cases} \\\\\\ A=10000\left(1+(0.0595)/(1)\right)^(1\cdot 10) \implies {\LARGE \begin{array}{llll} A \approx 17824.18 \end{array}}`