Question 446206: 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 stanbon(75887)   (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)
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At this point you should divide both sides by 450 to get:
2.2222... = (1.00417)^(12t)
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When the variable is in the exponent you need to take the
log to get it down where you can solve for it.
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You get:
(12t)*log(1.00417) = log(2.22222..)
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t = [log2.2222..]/[12log1.00417]
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t = 16 years
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Cheers,
Stan H.
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