MATH SOLVE

2 months ago

Q:
# Determine the time necessary for P dollars to double when it is invested at interest rate r compounded annually, monthly, daily, and continuously. (Round your answers to two decimal places.)r=14%

Accepted Solution

A:

Answer:Part 1) 8.17 yearsPart 2) 4.98 yearsPart 3) 4.95 yearsPart 4) 4.95 yearsStep-by-step explanation:we know that The compound interest formula is equal to [tex]A=P(1+\frac{r}{n})^{nt}[/tex] where A is the Final Investment Value P is the Principal amount of money to be invested r is the rate of interest in decimalt is Number of Time Periods n is the number of times interest is compounded per yearPart 1) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded annuallyin this problem we have [tex]t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=1[/tex] substitute in the formula above [tex]2p=p(1+\frac{0.14}{1})^{t}[/tex] [tex]2=(1.14)^{t}[/tex] Apply log both sides[tex]log(2)=log[(1.14)^{t}][/tex] [tex]log(2)=(t)log(1.14)[/tex] [tex]t=log(2)/log(1.14)[/tex] [tex]t=8.17\ years[/tex]Part 2) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded monthlyin this problem we have [tex]t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=12[/tex] substitute in the formula above [tex]2p=p(1+\frac{0.14}{12})^{12t}[/tex] [tex]2=(\frac{12.14}{12})^{12t}[/tex] Apply log both sides[tex]log(2)=log[(\frac{12.14}{12})^{12t}][/tex] [tex]log(2)=(12t)log(\frac{12.14}{12})[/tex] [tex]t=log(2)/12log(\frac{12.14}{12})[/tex] [tex]t=4.98\ years[/tex]Part 3) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded dailyin this problem we have [tex]t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=365[/tex] substitute in the formula above [tex]2p=p(1+\frac{0.14}{365})^{365t}[/tex] [tex]2=(\frac{365.14}{365})^{365t}[/tex] Apply log both sides[tex]log(2)=log[(\frac{365.14}{365})^{365t}][/tex] [tex]log(2)=(365t)log(\frac{365.14}{365})[/tex] [tex]t=log(2)/365log(\frac{365.14}{365})[/tex] [tex]t=4.95\ years[/tex]Part 4) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% continuouslywe know thatThe formula to calculate continuously compounded interest is equal to[tex]A=P(e)^{rt}[/tex] where A is the Final Investment Value P is the Principal amount of money to be invested r is the rate of interest in decimal t is Number of Time Periods e is the mathematical constant numberwe have [tex]t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14[/tex] substitute in the formula above [tex]2p=p(e)^{0.14t}[/tex] Simplify[tex]2=(e)^{0.14t}[/tex] Apply ln both sides[tex]ln(2)=ln[(e)^{0.14t}][/tex] [tex]ln(2)=(0.14t)ln(e)[/tex] Remember that ln(e)=1[tex]ln(2)=(0.14t)[/tex] [tex]t=ln(2)/(0.14)[/tex] [tex]t=4.95\ years[/tex]