Chapter ISG, Problem 7.5.4P

To determine

To find: The original investment of Graham’s, check this investment is increasing or decreasing and find the time when the investment will be worth half of its starting value.

Answer to Problem 7.5.4P

The original investment was $2400$ , the investment is decreasing with time and after about $13.5$ years the investment will be worth half of its starting value.

Explanation of Solution

Given information:

The equation is $y=2400{\left(0.95\right)}^x$ .

Calculation:

Substitute 0 for $x$ in the equation $y=2400{\left(0.95\right)}^x$ to find the original investment.

  $\begin{array}{c}y=2400{\left(0.95\right)}^0\\ =2400\left(1\right)\\ =2400\end{array}$

So, the original investment is $2400$ .

Find the value of $y$ for different values of $x$ to check the investment is increasing or decreasing.

Substitute $x=1$ in the equation $y=2400{\left(0.95\right)}^x$ .

  $\begin{array}{c}y=2400{\left(0.95\right)}^1\\ =2280\end{array}$

Substitute $x=2$ in the equation $y=2400{\left(0.95\right)}^x$ .

  $\begin{array}{c}y=2400{\left(0.95\right)}^2\\ =2166\end{array}$

So, as the value of $x$ increases the investment is decreasing.

Substitute $1200$ for $y$ in the equation $y=2400{\left(0.95\right)}^x$ to find the time when the investment will be worth half of its starting value.

  $\begin{array}{c}1200=2400{\left(0.95\right)}^x\\ {\left(0.95\right)}^x=\frac{1200}{2400}\\ {\left(0.95\right)}^x=0.5\end{array}$

Take the $\log$ on both sides of the equations.

  $\begin{array}{c}\log {\left(0.95\right)}^x=\log \left(0.5\right)\\ x\log \left(0.95\right)=\log \left(0.5\right)\\ x=\frac{\log \left(0.5\right)}{\log \left(0.95\right)}\\ \approx 13.5\text{Years}\end{array}$

Therefore, the original investment was $2400$ , the investment is decreasing with time and after about $13.5$ years the investment will be worth half of its starting value.