Chapter 7, Problem 63RE

a.

To determine

To Find:

To find the amount of time required for $\$10,000$ to double if $6.3\%$ of annual interest is compounded annually.

Answer to Problem 63RE

It takes $11.35$ years (or) $11$ years $4$ months $5$ days to double it up annually.

Explanation of Solution

Given information:

The given principal amount is $\$10,000$ and the rate of interest is $6.3\%$ .

Formula:

Compound interest formula is $A=P{\left(1+\frac{r}{n}\right)}^{nt}$ where $A$ -total amount; $P$ -principal amount;

  $r$ -rate of interest; $n$ -number of years; $t$ -time in years

The Compound interest formula is $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

From given we have $A=\$20,000;P=\$10,000;r=6.3\%;n=1$

Now substituting all the information in the formula,

  $\begin{array}{l}A=P{\left(1+\frac{r}{n}\right)}^{nt}\\ 20,000=10,000{\left(1+\frac{0.063}{1}\right)}^{1\cdot t}\\ 20,000=10,000{\left(1.063\right)}^{1\cdot t}\\ \frac{20,000}{10,000}={\left(1.063\right)}^{1\cdot t}\\ {\left(1.063\right)}^{1\cdot t}=2\end{array}$

Taking natural log on both sides,

  $\begin{array}{l}\ln \left({1.063}^{1\cdot t}\right)=\ln \left(2\right)\\ 1\cdot t\cdot \ln \left(1.063\right)=\ln \left(2\right)\\ 1\cdot t=\frac{\ln \left(2\right)}{\ln \left(1.063\right)}\\ 1\cdot t=11.34538\\ t=11.35\end{array}$

Therefore, it takes $11.35$ years (or) $11$ years $4$ months $5$ days to double it up annually.

b.

To determine

To Estimate:

To find the amount of time required for $\$10,000$ to double if $6.3\%$ of annual interest is compounded continuously.

Answer to Problem 63RE

It takes $11$ years to double it up.

Explanation of Solution

Given information:

The given principal amount is $\$10,000$ and the rate of interest is $6.3\%$ .

Formula:

The amount of money in the account after t years is

  $A\left(t\right)=A_0{e}^{rt}$

Here $A_0=10,000$ and $r=6.3\%$

  $A\left(t\right)=20,000$

Now substituting all the information in the formula,

  $\begin{array}{l}A\left(t\right)=A_0{e}^{rt}\\ 20,000=10,000e^{0.063t}\\ \frac{20,000}{10,000}=e^{0.063t}\\ 2=e^{0.063t}\end{array}$

Taking natural log on both sides,

  $\begin{array}{l}\ln \left(2\right)=\ln \left(e^{0.063t}\right)\\ \ln \left(2\right)=0.063t\ln \left(e\right)\\ \ln \left(2\right)=0.063t\\ \frac{\text{0}\text{.69314}}{0.063}=t\\ t=11.00158\end{array}$

Therefore, it takes $11$ years to double it up continuously.