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Rearranging an equation using log functions

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi guys. I'm trying to rearrage an equation using the natural log and I thought I had it correct but am now struggling. I need to rearange it into the form: y=a0 + a1*C. Where a0 and a1 are some combination of the other variables and numbers below

    2. Relevant equations

    H = ((2^(X-1))-1)/(K*(X-1)*C^(X-1))

    H,X,K and C are all variables

    3. The attempt at a solution

    I've tried rearanging using log laws but I now think it might be wrong.


    ln(H) = ln(2^(X-1)-1) - ln(K*(X-1)) - (X-1)*ln(C)

    But I'm not sure if this is legal considering the equation could be rearranged to:

    H = (2^(X-1))/(K*(X-1)*C^(X-1)) - 1/(K*(X-1)*C^(X-1))

    Can I then apply log as follows:

    ln(H) = ln(2^(X-1))/(K*(X-1)*C^(X-1)) - ln(1/(K*(X-1)*C^(X-1)))

    or can I only apply log once to each side, eg:

    ln(H) = ln((2^(X-1))/(K*(X-1)*C^(X-1))- 1/(K*(X-1)*C^(X-1)))
     
  2. jcsd
  3. Oct 27, 2011 #2

    BruceW

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    Homework Helper

    ln(H) = ln(2^(X-1)-1) - ln(K*(X-1)) - (X-1)*ln(C)
    H = (2^(X-1))/(K*(X-1)*C^(X-1)) - 1/(K*(X-1)*C^(X-1))
    These two are fine. But
    ln(H) = ln(2^(X-1))/(K*(X-1)*C^(X-1)) - ln(1/(K*(X-1)*C^(X-1)))
    Is not right. You can only apply log once to each side (as you said), it would be:
    ln(H) = ln((2^(X-1))/(K*(X-1)*C^(X-1))- 1/(K*(X-1)*C^(X-1)))

    But What is it you are trying to do? Get C to be the subject of the equation?
     
  4. Oct 27, 2011 #3
    I was trying to rearrange it to the form:

    y= a0 +a1*C

    or

    ln(y) = a0 + a1*ln(C)

    I was trying to do this so a program called polymath could determine the a0 and a1 (using experimental date), so that I could find the parameters X and K for the given set of data.

    I have since worked it out though. Thank you for your help anyway though.
     
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