Chapter 3.5, Problem 16E

(a)

To determine

To find : the time necessary for P dollars to double (when investment compounded annually).

Answer to Problem 16E

The time necessary for P dollars to double(Compounded annually) is $\underset{\_}{t\approx 11years}$

Explanation of Solution

Given information : Amount invested is P dollars, annual rate of interest is 6.5%, 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 = 0.065, & n = 1 because compounded annually}	$2P=P{\left(1+\frac{0.065}{1}\right)}^{1\cdot t}$
Use symmetric property of equality which states that if a = b then b = a to rewrite the equation	$P{\left(1+\frac{0.065}{1}\right)}^{1\cdot t}=2P$

Calculation (Continued):

Description	Steps
Simplify the expression in the left side of the equation	$P{\left(1.065\right)}^t=2P$
Divide by $P$on both sides	$\frac{P{\left(1.065\right)}^t}{P}=\frac{2P}{P}$
Simplifying fraction on both sides	${\left(1.065\right)}^t=2$
Take natural logarithm on both sides	$\ln {\left(1.065\right)}^t=\ln \left(2\right)$
Apply the logarithmic rule $\ln m^n=n\ln m$ in left side of the equation	$t\cdot \ln \left(1.065\right)=\ln 2$
Divide by ln(1.065) on both sides	$\frac{t\cdot \ln 1.065}{\ln 1.065}=\frac{\ln 2}{\ln 1.065}$
Simplify fraction on both sides of the equation	$t\approx 11years$

Conclusion:

It would take time of 11years for P dollars to double when it is invested at interest rate $r=6.5\%$ compounded annually.

(b)

To determine

To find : the time necessary for P dollars to double when investment compounded monthly.

Answer to Problem 16E

The time necessary for P dollars to double(compounded monthly) is $\underset{\_}{t\approx 10.6927years}$

Explanation of Solution

Given information : Amount invested is P dollars, annual rate of interest is 6.5%, 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 = 0.065, & n = 12 because compounded monthly}	$2P=P{\left(1+\frac{0.065}{12}\right)}^{12\cdot t}$
Use symmetric property of equality which states that if a = b then b = a to rewrite the equation	$P{\left(1+\frac{0.065}{12}\right)}^{12\cdot t}=2P$

Calculation (Continued):

Description	Steps
Simplify the expression in the left side of the equation	$P{\left(1.00541\overline{6}\right)}^{12t}=2P$
Divide by $P$on both sides	$\frac{P{\left(1.00541\overline{6}\right)}^{12t}}{P}=\frac{2P}{P}$
Simplifying fraction on both sides	${\left(1.00541\overline{6}\right)}^{12t}=2$
Take natural logarithm on both sides	$\ln {\left(1.00541\overline{6}\right)}^{12t}=\ln \left(2\right)$
Apply the logarithmic rule $\ln m^n=n\ln m$ in left side of the equation	$12t\cdot \ln \left(1.00541\overline{6}\right)=\ln 2$
Divide by $12\ln \left(1.00541\overline{6}\right)$on both sides	$\frac{t\cdot 12\ln \left(1.00541\overline{6}\right)}{12\ln \left(1.00541\overline{6}\right)}=\frac{\ln 2}{12\ln \left(1.00541\overline{6}\right)}$
Simplify fraction on both sides of the equation	$t\approx 10.6927years$

Conclusion:

It would take time of 10.6927 years for P dollars to double when it is invested at interest rate $r=6.5\%$ compounded monthly.

(c)

To determine

To find : the time necessary for P dollars to double when investment compounded daily.

Answer to Problem 16E

The time necessary for P dollars to double (compounded daily) is $\underset{\_}{t\approx 10.6648years}$

Explanation of Solution

Given information : Amount invested is P dollars, annual rate of interest is 6.5%, 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 = 0.065, & n = 365 because compounded daily}	$2P=P{\left(1+\frac{0.065}{365}\right)}^{365\cdot t}$
Use symmetric property of equality which states that if a = b then b = a to rewrite the equation	$P{\left(1+\frac{0.065}{365}\right)}^{365\cdot t}=2P$

Calculation (Continued):

Description	Steps
Simplify the expression in the left side of the equation	$P{\left(1+\frac{0.065}{365}\right)}^{365\cdot t}=2P$
Divide by $P$on both sides	$\frac{P{\left(1+\frac{0.065}{365}\right)}^{365t}}{P}=\frac{2P}{P}$
Simplifying fraction on both sides	${\left(1+\frac{0.065}{365}\right)}^{365t}=2$
Take natural logarithm on both sides	$\ln {\left(1+\frac{0.065}{365}\right)}^{365t}=\ln \left(2\right)$
Apply the logarithmic rule $\ln m^n=n\ln m$ in left side of the equation	$365\ln \left(1+\frac{0.065}{365}\right)\cdot t=\ln 2$
Divide by $365\ln \left(1+\frac{0.065}{365}\right)$on both sides	$\frac{365\ln \left(1+\frac{0.065}{365}\right)\cdot t}{365\ln \left(1+\frac{0.065}{365}\right)}=\frac{\ln 2}{365\ln \left(1+\frac{0.065}{365}\right)}$
Simplify fraction on both sides of the equation	$t\approx 10.6648years$

Conclusion:

It would take time of 10.6648 years for P dollars to double when it is invested at interest rate $r=6.5\%$ compounded daily.

(d)

To determine

To find : the time necessary for P dollars to double when investment compounded continuously.

Answer to Problem 16E

The time necessary for P dollars to double (compounded continuously) is $\underset{\_}{t\approx 6.6638years}$

Explanation of Solution

Given information : Amount invested is P dollars, annual rate of interest is 6.5%, 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 continuous 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: $A=P{e}^{rt}$

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

Calculation:

Description	Steps
Substitute the given information in the formula $A=P{e}^{rt}$because amount compounding continuously{A=2P because amount doubling, r = 0.065}	$2P=P{e}^{0.065\cdot t}$
Use symmetric property of equality which states that if a = b then b = a to rewrite the equation	$P{e}^{0.065\cdot t}=2P$

Calculation (Continued):

Description	Steps
Divide by $P$on both sides	$\frac{P{e}^{0.065\cdot t}}{P}=\frac{2P}{P}$
Simplifying fraction on both sides	$e^{0.065\cdot t}=2$
Take natural logarithm on both sides	$\ln e^{0.065\cdot t}=\ln \left(2\right)$
Apply the logarithmic rule $\ln e^X=X$ in left side of the equation	$0.065t=\ln 2$
Divide by 0.065 on both sides	$\frac{0.065t}{0.065}=\frac{\ln 2}{0.065}$
Simplify fraction on both sides of the equation	$t\approx 10.6638years$

Conclusion:

It would take time of 10.6638 years for P dollars to double when it is invested at interest rate $r=6.5\%$ compounded continuously.