Chapter 5.7, Problem 39AYU

Time Required to Reach a Goal If Tanisha has $\$100$ to invest at $4\%$ per annum compounded monthly, how long will it be before she has $\$150$? If the compounding is continuous, how long will it be?

To determine

To find: Time Required to Reach a Goal If Tanisha has $\text{\$100}$ to invest at $\text{4\%}$ per annum compounded monthly, how long will it be before she has $\text{\$150}$? If the compounding is continuous, how long will it be?

Answer to Problem 39AYU

Solution:

a. $10.15$ years long will it be before she has $\text{\$150}$ compounded monthly.

b. $10.14$ years long will be if the compounding is continuous.

Explanation of Solution

Given:

$P=100, r=4\%=0.04,$ and $A=150$

Calculation:

a. $P=100, r=4\%=0.04, n=12$ and $A=150$

$A=P {\left(1+\frac{r}{n}\right)}^{nt}$

$150=100 {\left(1+\frac{0.04}{12}\right)}^{12t}$

$\frac{150}{100}={\left(1+\frac{0.04}{12}\right)}^{12t}$

$1.5={\left(1+\frac{0.04}{12}\right)}^{12t}$

Taking $\text{log}$ on both sides

$\log 1.5=\log {\left(1+\frac{0.04}{12}\right)}^{12t}$

$\log 1.5=12t\mathrm{ log}\left(1+\frac{0.04}{12}\right)$

$12t=\frac{\log 1.5}{\log \left(1+\frac{0.04}{12}\right)}$

$12t=121.84$

$t=10.15$ years

b. $A=P e^{rt}$

$150=100e^{0.04t}$

$\frac{150}{100}=e^{0.04t}$

$e^{0.04t}=1.5$

Taking $\text{log}$ on both sides

$\ln e^{0.04t}=\ln 1.5$

$0.04 t=\ln 1.5$

$t=\frac{\ln 1.5}{0.04}=10.14$ years