Question

How long will it take for $2500 to double if it is invested at 6% annual interest compounded 6 times a year? Enter exact calculations or round to 3 decimal places.
- It will take ____years to double.
How long will it take if the interest is compounded continuously?
-compounded continuously, it would only take ____years

Answer 1

Answer:

Part 1) $t=11.610\ years$

Part 2) $t=11.552\ years$

Step-by-step explanation:

Part 1) 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

in this problem we have  

$A=\ 5,000\\P=\ 2,500\\ r=6\%=6/100=0.06\\n=6$  

substitute in the formula above

$5,000=2,500(1+\frac{0.06}{6})^{6t}$  

$2=(1.01)^{6t}$  

Apply log both sides

$log(2)=log[(1.01)^{6t}]$

$log(2)=(6t)log(1.01)$  

solve for t

$t=log(2)/[6log(1.01)]$  

$t=11.610\ years$

Part 2) 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  

$A=\ 5,000\\P=\ 2,500\\ r=6\%=6/100=0.06$  

substitute in the formula above

$5,000=2,500(e)^{0.06t}$

$2=(e)^{0.06t}$  

Apply ln both sides

$ln(2)=ln[(e)^{0.06t}]$

$ln(2)=(0.06t)ln(e)$

$ln(2)=(0.06t)$

$t=ln(2)/(0.06)$

$t=11.552\ years$