Chapter 8.7, Problem 6BCYP

To determine

Find the interest rate.

Answer to Problem 6BCYP

  $2.6\mathrm{\%}$ .

Explanation of Solution

Given information:

The investment of $\mathrm{\$}10,000$ , the account need to grow $\mathrm{\$}30,000$ in $18\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}.$ .

Calculation:

Consider, the formula for interest compounded continuously.

  $\mathrm{A}=\mathrm{P}{\mathrm{e}}^{\mathrm{r}\mathrm{t}}$

  $\begin{array}{l}\mathrm{P}=\mathrm{\$}10,000,\mathrm{A}=30,000,\mathrm{t}=18\\ 30000=10000{\mathrm{e}}^{\left(\mathrm{r}\right)18}\\ 30000/10000={\mathrm{e}}^{\left(\mathrm{r}\right)18}\\ 3={\mathrm{e}}^{\left(\mathrm{r}\right)18}\\ \mathrm{t}\mathrm{a}\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{g}\log \mathrm{b}\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}\\ \log 3=\log {\mathrm{e}}^{\left(\mathrm{r}\right)18}\left[\log {\mathrm{e}}^{\mathrm{x}}=\mathrm{x}\right]\\ \log 3=\mathrm{r}18\\ 0.477/18=\mathrm{r}\\ \mathrm{r}=0.026\mathrm{o}\mathrm{r}\approx 2.6\mathrm{\%}\end{array}$

Hence, the interest rate $2.6\mathrm{\%}$ .