Answer :
Answer:
Part 1) 8.17 years
Part 2) 4.98 years
Part 3) 4.95 years
Part 4) 4.95 years
Step-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 decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
Part 1) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded annually
in 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 monthly
in 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 daily
in 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% continuously
we know that
The 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 number
we 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]