Question

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%

Answer 1

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  

$A=P(1+\frac{r}{n})^{nt}$  

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  

$t=?\ years\\ P=\ p\\A=\ 2p\\r=14\%=14/100=0.14\\n=1$  

substitute in the formula above  

$2p=p(1+\frac{0.14}{1})^{t}$  

$2=(1.14)^{t}$  

Apply log both sides

$log(2)=log[(1.14)^{t}]$  

$log(2)=(t)log(1.14)$  

$t=log(2)/log(1.14)$  

$t=8.17\ years$

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      

$t=?\ years\\ P=\ p\\A=\ 2p\\r=14\%=14/100=0.14\\n=12$  

substitute in the formula above  

$2p=p(1+\frac{0.14}{12})^{12t}$  

$2=(\frac{12.14}{12})^{12t}$  

Apply log both sides

$log(2)=log[(\frac{12.14}{12})^{12t}]$  

$log(2)=(12t)log(\frac{12.14}{12})$  

$t=log(2)/12log(\frac{12.14}{12})$  

$t=4.98\ years$

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  

$t=?\ years\\ P=\ p\\A=\ 2p\\r=14\%=14/100=0.14\\n=365$  

substitute in the formula above  

$2p=p(1+\frac{0.14}{365})^{365t}$  

$2=(\frac{365.14}{365})^{365t}$  

Apply log both sides

$log(2)=log[(\frac{365.14}{365})^{365t}]$  

$log(2)=(365t)log(\frac{365.14}{365})$  

$t=log(2)/365log(\frac{365.14}{365})$  

$t=4.95\ years$

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

$A=P(e)^{rt}$  

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  

$t=?\ years\\ P=\ p\\A=\ 2p\\r=14\%=14/100=0.14$  

substitute in the formula above  

$2p=p(e)^{0.14t}$  

Simplify

$2=(e)^{0.14t}$  

Apply ln both sides

$ln(2)=ln[(e)^{0.14t}]$  

$ln(2)=(0.14t)ln(e)$  

Remember that ln(e)=1

$ln(2)=(0.14t)$  

$t=ln(2)/(0.14)$  

$t=4.95\ years$