Question

At an interest rate of 8% compounded annually, how long will it take to double the following investments?

Answer 1

Let's see, if the first one has a Principal of $50, when it doubles the accumulated amount will then be $100,

recall your logarithm rules for an exponential,

$\bf \textit{Logarithm of exponentials}\\\\ log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\ -------------------------------\\\\ \qquad \textit{Compound Interest Earned Amount}$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\to &\ 100\\ P=\textit{original amount deposited}\to &\ 50\\ r=rate\to 8\%\to \frac{8}{100}\to &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annnually, thus once} \end{array}\to &1\\ t=years \end{cases} \\\\\\ 100=50\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 100=50(1.08)^t \\\\\\ \cfrac{100}{50}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)$

$\bf log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\ -------------------------------$

now, for the second amount, if the Principal is 500, the accumulated amount is 1000 when doubled,

$\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\to &\ 1000\\ P=\textit{original amount deposited}\to &\ 500\\ r=rate\to 8\%\to \frac{8}{100}\to &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annnually, thus once} \end{array}\to &1\\ t=years \end{cases} \\\\\\ 1000=500\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 1000=500(1.08)^t$

$\bf \cfrac{1000}{500}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t) \\\\\\ log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\ -------------------------------$

now, for the last, Principal is 1700, amount is then 3400,

$\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\to &\ 3400\\ P=\textit{original amount deposited}\to &\ 1700\\ r=rate\to 8\%\to \frac{8}{100}\to &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annnually, thus once} \end{array}\to &1\\ t=years \end{cases}$

$\bf 3400=1700\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 3400=1700(1.08)^t \\\\\\ \cfrac{3400}{1700}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t) \\\\\\ log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t$