Question

$750, annual rate:10 1/2% what is the time to double & the amount after 10 years?

Answer 1

well, let's start like the crab, from the back, what would the amount be after 10 years, with a 10.5% annual rate, compounding annually of course.

`~~~~~~ \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 &\$750\\ r=rate\to 10.5\%\to (10.5)/(100)\dotfill &0.105\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &10 \end{cases} \\\\\\ A=750\left(1+(0.105)/(1)\right)^(1\cdot 10) \implies A=750(1.105)^(10)\implies \boxed{A \approx 2035.56}`

when will it double? well, hmmm 750 doubled will be 1500, so

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$1500\\ P=\textit{original amount deposited}\dotfill &\$750\\ r=rate\to 10.5\%\to (10.5)/(100)\dotfill &0.105\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years \end{cases}`

`1500=750\left(1+(0.105)/(1)\right)^(1\cdot t) \implies \cfrac{1500}{750}=(1.105)^t\implies 2=1.105^t \\\\\\ \log(2)=\log(1.105^t)\implies \log(2)=t\log(1.105) \\\\\\ \cfrac{\log(2)}{\log(1.105)}=t\implies 6.94\approx t\qquad \textit{about 6 years and 344 days}`