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Exponential Variable and Logarithms

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find the value of a variable which happens to be an exponent.

    2. Relevant equations

    [itex]230,000=1500\frac{(1+.00077)^n-1}{.00077}[/itex]

    I believe I need to use logarithms to get to my answer, but I've reviewed logarithm rules and I'm stuck.

    3. The attempt at a solution

    I've divided 230k by 1500, but am stuck at that point.

    I'm now at

    [itex]153.33=\frac{(1+.00077)^n-1}{.00077}[/itex]

    ...and stuck :( I think I need to log both sides, something like [itex]log 153.33=(n)log \frac{(1+.00077)-1}{.00077}[/itex]
    I've tried simplifying the right side, but I end up at 1 which means the right side ends up being zero, which doesn't make any sense.

    Any assistance would be greatly appreciated.
     
  2. jcsd
  3. Mar 5, 2013 #2

    Fredrik

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    Do you know how to solve ##x^n=y## for n? Can you rewrite the equation in that form?
     
  4. Mar 5, 2013 #3

    eumyang

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    No... you can't do that! Before taking the logarithm of both sides, I would isolate the (1+.00077)^n portion. Can you do that?
     
  5. Mar 5, 2013 #4
    Would it be [itex]n=\frac{log153.33}{log(1+.00077)-1}{.00077}[/itex]
     
    Last edited: Mar 5, 2013
  6. Mar 5, 2013 #5
    Yeah, nevermind that can't be right. Very frustrating..
     
  7. Mar 5, 2013 #6
    I mean, I know that if x^n=y then ln(y)/ln(x)=n. I can't figure out how to write using my values though.
     
  8. Mar 5, 2013 #7

    SteamKing

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    No. In addition to reviewing the rules about logarithms, you need to brush up on your algebra as well.

    You had 153.33 = ((1+0.00077)^n - 1) / 0.00077

    You can further simplify:

    153.33(0.00077) = (1+0.00077)^n - 1
    153.33(0.00077)+1 = (1+0.00077)^n

    Now use logarithms:

    log (153.33(0.00077) + 1) = n log (1+0.00077)

    Therefore:

    n = log(153.33(0.00077)+1) / log (1+0.00077)

    or

    n = 144.99
     
  9. Mar 5, 2013 #8

    Fredrik

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    You probably shouldn't give away the complete solution like that. A better hint would be to ask if jsully knows how to solve
    $$a=b\frac{x-1}{c}$$ for x when a,b,c are non-zero real numbers.
     
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