Chapter 3.5, Problem 7E

To determine

To find : the time to double and amount after 10 years.

Answer to Problem 7E

The time to double is $\underset{\_}{t\approx 19.8years}$

The amount after 10 years is $\underset{\_}{A\approx \$1419.07}$

Explanation of Solution

Given information : Initial Investment is $1000 & Annual percentage Rate is 3.5%

Concept Involved:

Solving for a variable means getting the variable alone in one side of the equation by undoing whatever operation is done to it.

Formula Used:

For continuous compounding, after t years, the balance A in an account with principal P and annual interest rate r (in decimal form) is given by the formula: $A=P{e}^{rt}$

Logarithmic property: $\ln e^X=X$

Calculation:

Description	Steps
Substitute the given information in the formula	$A=1000e^{0.035t}$
To find the time to double, we need to replace 2000 in place of A	$2000=1000e^{0.035t}$
Use symmetric property of equality which states that if a = b then b = a to rewrite the equation	$1000e^{0.035t}=2000$
Divide by 1000 on both sides	$\frac{1000e^{0.035t}}{1000}=\frac{2000}{1000}$
Simplify fraction on both sides of the equation	$e^{0.035t}=2$
Take natural logarithm on both sides	$\ln e^{0.035t}=\ln \left(2\right)$
Apply the logarithmic rule $\ln e^X=X$ in left side of the equation	$0.035t=\ln 2$

Calculation (Continued):

Description	Steps
Divide by 0.035 on both sides	$\frac{0.035t}{0.035}=\frac{\ln 2}{0.035}$
Simplify fraction on both sides of the equation	$t=19.8years$
To find the amount after 10 years substitute 10 for t in the equation $A=1000e^{0.035t}$	$\begin{array}{l}A=1000e^{0.035\left(10\right)}\\ =1000e^{0.35}\\ \approx \$1419.07\end{array}$

Conclusion:

Initial Investment	Annual % Rate	Time to Double	Amount After 10 years
$1000 (Given)	3.5% (Given)	19.8 years	$1419.07