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Question 445137: The question is: The amount of money A accrued at the end of t years when a certain amount, P, is invested at an annual rate r compounded n times per year is given by: A=P(1+r/n)^nt. How long will it take for $450 at 5% interest compounded monthly to reach $1000? (Write your answer to two decimal places.)
I've set the problem up like this:
1000=450(1+.05/12)^12(t)
1000=450(1+0.00417)^12(t)
1000=450(1.00417)^12(t)
1000=450(1.0512)^t
1000=473.04^t
And that's where I'm stuck. Assuming my math to that point is accurate (and if it's not, could you please help?), how do I go about the problem from this point? How do I get t to no longer be an exponent and move on to get the time?
Thank you!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! I've set the problem up like this:
1000=450(1+.05/12)^12(t)
1000=450(1+0.00417)^12(t)
1000=450(1.00417)^12(t)
1000=450(1.0512)^t
1000=473.04^t
..
You are correct up to the third step.
1000=450(1.00417)^12(t)
divide both sides by 450
1000/450=(1.00417)^12(t)
Take the logarithms of both sides
log(1000/450)=12t*log(1.00417)
12t=log(1000/450)/log(1.00417)=191.8876
t=191.8876/12=15.99 years (ans)
note: Usually, when the unknown is an exponent logs are required to solve.
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