Question

$3000 is invested in an account at interest rate r, compounded continuously. Find the time required for the amount to triple. (Approximate the result to two decimal places.)

Answer 1

Answer:

$36.62\ years$  

Step-by-step explanation:

The value of r is the question is

r=0.03

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  

$t=?\ years\\ P=\ 3,000\\A=\ 9,000\\r=0.03$  

substitute in the formula above

$9,000=3,000(e)^{0.03t}$  

solve for t

$3=(e)^{0.03t}$  

Apply ln both sides

$ln(3)=ln[(e)^{0.03t}]$  

Apply property of logarithms

$ln(3)=(0.03t)ln[(e)]$  

Remember that

$ln(e)=1$

$ln(3)=(0.03t)$  

$t=ln(3)/(0.03)=36.62\ years$