Chapter 5.7, Problem 40AYU

Time Required to Reach a Goal If Angela has $\$100$ to invest at $2.5\%$ per annum compounded monthly, how long will it be before she has $\$175$? If the compounding is continuous, how long will it be?

To determine

To find: Time Required to Reach a Goal If Angela has $\text{\$100}$ to invest at $\text{2}\text{.5\%}$ per annum compounded monthly, how long will it be before she has $\text{\$175?}$ If the compounding is continuous, how long will it be?

Answer to Problem 40AYU

Solution:

a. $t = 22.41$ years.

b. $t = 22.39$ years.

Explanation of Solution

Given:

$P=\mathrm{\$100}, r=2.5\%=0.025,$ and $A=175$

Calculation:

a. $P=100, r=2.5\%=0.025, n=12$ and $A=175$

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

$175=100{\left(1+\frac{0.025}{12}\right)}^{12t}$

$\frac{175}{100}={\left(1+\frac{0.025}{12}\right)}^{12t}$

$1.75={\left(1+\frac{0.025}{12}\right)}^{12t}$

Taking $\text{log}$ on both sides

$\log 1.75=\log {\left(1+\frac{0.025}{12}\right)}^{12t}$

$\log 1.75=12t\mathrm{ log}\left(1+\frac{0.025}{12}\right)$

$12t=\frac{\log 1.75}{\log \left(1+\frac{0.025}{12}\right)}$

$12t=268.895$

$t=22.41$ years

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

$175=100e^{0.025t}$

$\frac{175}{100}=e^{0.025t}$

$e^{0.025t}=1.75$

Taking $\text{log}$ on both sides

$\ln e^{0.025t}=\ln 1.75$

$0.025t=\ln 1.75$

$t=\frac{\ln 1.75}{0.025}=10.14$

$t=22.39$ years.