SOLUTION: Find the time required for an investment of 5000 dollars to grow to 7200 dollars at an interest rate of 7.5 percent per year, compounded quarterly.

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Question 662331: Find the time required for an investment of 5000 dollars to grow to 7200 dollars at an interest rate of 7.5 percent per year, compounded quarterly.
Found 2 solutions by sravanik2005, lwsshak3:
Answer by sravanik2005(14) About Me  (Show Source):
You can put this solution on YOUR website!
A = P(1+ r/n)^nt
here P is principal amount=5000
r=interest rate as decimal=0.075
and n=number of times the interest is compounded per year=4
t= number of years=? and A=amount of money accumulated after t years, including interest=7200 dollars
substitute these values in that formula
7200=5000(1+0.075/4)^(4t)
7200/5000=1.01875^4t => 1.01875^4t=36/25
ln(1.01875)^4t=ln (36/25)
4t(ln 1.01875)=ln(36/25)
=>t= ln(36/25)/(4ln 1.01875)=4.91 that is approximately 5 years

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the time required for an investment of 5000 dollars to grow to 7200 dollars at an interest rate of 7.5 percent per year, compounded quarterly.
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compound interest formula: A=P(1+r)^n, P=initial investment, r=interest rate/period, n=number of periods, A=amt after n-periods.
For given problem:
P=5000
r=.075/4=.01875
n=to find
A=7200
..
7200=5000(1+.01875)^n
7200/5000=(1.0175)^n
1.44=(1.0175)^n
take log of both sides
log(1.44)=n*log(1.0175)
n=log1.44/log(1.0175)
n≈21 quarters or 5.25 years