Chapter 3.5, Problem 15E

(a)

To determine

To find : the time necessary for P dollars to double.

Answer to Problem 15E

The time necessary for P dollars to double is $\underset{\_}{t\approx 7.2725years}$

Explanation of Solution

Given information : Amount invested is P dollars, annual rate of interest is 10%, and it compounded annually

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 periodic compounding, after t years, the balance A in an account with principal P, number of times interest applied per time period n and annual interest rate r (in decimal form) is given by the formula: $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Logarithmic property: $\ln m^n=n\ln m$

Calculation:

Description	Steps
Substitute the given information in the formula $A=P{\left(1+\frac{r}{n}\right)}^{nt}${A=2P because amount doubling, r = 0.1, & n = 1 because compounded annually}	$2P=P{\left(1+\frac{0.1}{1}\right)}^{1\cdot t}$
Use symmetric property of equality which states that if a = b then b = a to rewrite the equation	$P{\left(1+\frac{0.1}{1}\right)}^{1\cdot t}=2P$

Calculation (Continued):

Description	Steps
Simplify the expression in the left side of the equation	$P{\left(1.1\right)}^t=2P$
Divide by $P$on both sides	$\frac{P{\left(1.1\right)}^t}{P}=\frac{2P}{P}$
Simplifying fraction on both sides	${\left(1.1\right)}^t=2$
Take natural logarithm on both sides	$\ln {\left(1.1\right)}^t=\ln \left(2\right)$
Apply the logarithmic rule $\ln m^n=n\ln m$ in left side of the equation	$t\cdot \ln \left(1.1\right)=\ln 2$
Divide by ln(1.1) on both sides	$\frac{t\cdot \ln 1.1}{\ln 1.1}=\frac{\ln 2}{\ln 1.1}$
Simplify fraction on both sides of the equation	$t\approx 7.2725years$

Conclusion:

It would take time of 7.2725years for P dollars to double when it is invested at interest rate $r=10\%$ compounded annually.

(b)

To determine

To find : the time necessary for P dollars to double.

Answer to Problem 15E

The time necessary for P dollars to double is $\underset{\_}{t\approx 6.9603years}$

Explanation of Solution

Given information : Amount invested is P dollars, annual rate of interest is 10%, and it compounded monthly

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 periodic compounding, after t years, the balance A in an account with principal P, number of times interest applied per time period n and annual interest rate r (in decimal form) is given by the formula: $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Logarithmic property: $\ln m^n=n\ln m$

Calculation:

Description	Steps
Substitute the given information in the formula $A=P{\left(1+\frac{r}{n}\right)}^{nt}${A=2P because amount doubling, r = 0.1, & n = 12 because compounded monthly}	$2P=P{\left(1+\frac{0.1}{12}\right)}^{12\cdot t}$
Use symmetric property of equality which states that if a = b then b = a to rewrite the equation	$P{\left(1+\frac{0.1}{12}\right)}^{12\cdot t}=2P$

Calculation (Continued):

Description	Steps
Simplify the expression in the left side of the equation	$P{\left(1.008\overline{3}\right)}^{12t}=2P$
Divide by $P$on both sides	$\frac{P{\left(1.008\overline{3}\right)}^{12t}}{P}=\frac{2P}{P}$
Simplifying fraction on both sides	${\left(1.008\overline{3}\right)}^{12t}=2$
Take natural logarithm on both sides	$\ln {\left(1.008\overline{3}\right)}^{12t}=\ln \left(2\right)$
Apply the logarithmic rule $\ln m^n=n\ln m$ in left side of the equation	$12t\cdot \ln \left(1.008\overline{3}\right)=\ln 2$
Divide by $12\ln \left(1.008\overline{3}\right)$on both sides	$\frac{t\cdot 12\ln \left(1.008\overline{3}\right)}{12\ln \left(1.008\overline{3}\right)}=\frac{\ln 2}{12\ln \left(1.008\overline{3}\right)}$
Simplify fraction on both sides of the equation	$t\approx 6.9603years$

Conclusion:

It would take time of 6.9603 years for P dollars to double when it is invested at interest rate $r=10\%$ compounded annually.

(c)

To determine

To find : the time necessary for P dollars to double.

Answer to Problem 15E

The time necessary for P dollars to double is $\underset{\_}{t\approx 6.9324years}$

Explanation of Solution

Given information : Amount invested is P dollars, annual rate of interest is 10%, and it compounded daily

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 periodic compounding, after t years, the balance A in an account with principal P, number of times interest applied per time period n and annual interest rate r (in decimal form) is given by the formula: $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Logarithmic property: $\ln m^n=n\ln m$

Calculation:

Description	Steps
Substitute the given information in the formula $A=P{\left(1+\frac{r}{n}\right)}^{nt}$ {A=2P because amount doubling, r = 0.1, & n = 365 because compounded daily}	$2P=P{\left(1+\frac{0.1}{365}\right)}^{365\cdot t}$
Use symmetric property of equality which states that if a = b then b = a to rewrite the equation	$P{\left(1+\frac{0.1}{365}\right)}^{365\cdot t}=2P$

Calculation (Continued):

Description	Steps
Simplify the expression in the left side of the equation	$P{\left(1+\frac{0.1}{365}\right)}^{365\cdot t}=2P$
Divide by $P$on both sides	$\frac{P{\left(1+\frac{0.1}{365}\right)}^{365t}}{P}=\frac{2P}{P}$
Simplifying fraction on both sides	${\left(1+\frac{0.1}{365}\right)}^{365t}=2$
Take natural logarithm on both sides	$\ln {\left(1+\frac{0.1}{365}\right)}^{365t}=\ln \left(2\right)$
Apply the logarithmic rule $\ln m^n=n\ln m$ in left side of the equation	$365\ln \left(1+\frac{0.1}{365}\right)\cdot t=\ln 2$
Divide by   $365\ln \left(1+\frac{0.1}{365}\right)$on both sides	$\frac{365\ln \left(1+\frac{0.1}{365}\right)\cdot t}{365\ln \left(1+\frac{0.1}{365}\right)}=\frac{\ln 2}{365\ln \left(1+\frac{0.1}{365}\right)}$
Simplify fraction on both sides of the equation	$t\approx 6.9324years$

Conclusion:

It would take time of 6.9324 years for P dollars to double when it is invested at interest rate $r=10\%$ compounded annually.

(d)

To determine

To find : the time necessary for P dollars to double.

Answer to Problem 15E

The time necessary for P dollars to double is $\underset{\_}{t\approx 6.9314years}$

Explanation of Solution

Given information : Amount invested is P dollars, annual rate of interest is 10%, and it compounded continuously

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 periodic compounding, after t years, the balance A in an account with principal P, number of times interest applied per time period n and annual interest rate r (in decimal form) is given by the formula: $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

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

Calculation:

Description	Steps
Substitute the given information in the formula $A=P{\left(1+\frac{r}{n}\right)}^{nt}${A=2P because amount doubling, r = 0.1, & n = 12 because compounded continuously}	$2P=P{e}^{0.1\cdot t}$
Use symmetric property of equality which states that if a = b then b = a to rewrite the equation	$P{e}^{0.1\cdot t}=2P$

Calculation (Continued):

Description	Steps
Divide by $P$on both sides	$\frac{P{e}^{0.1\cdot t}}{P}=\frac{2P}{P}$
Simplifying fraction on both sides	$e^{0.1t}=2$
Take natural logarithm on both sides	$\ln e^{0.1t}=\ln \left(2\right)$
Apply the logarithmic rule $\ln e^X=X$ in left side of the equation	$0.1t=\ln 2$
Divide by 12 on both sides	$\frac{0.1t}{0.1}=\frac{\ln 2}{0.1}$
Simplify fraction on both sides of the equation	$t\approx 6.9314years$

Conclusion:

It would take time of 6.9314 years for P dollars to double when it is invested at interest rate $r=10\%$ compounded continuously.