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Having trouble with this log equation `

  1. Jan 1, 2007 #1
    How would one go about solving for C in the following equation?

    log ( A - B - C ) = -C + A + B

    I have done this...

    log ( A - B ) / log ( C ) = -C + A + B

    Then...

    C + ( log ( A - B ) / log ( C )) = A + B

    But I am just shooting in the dark really. Where would one go from here? I would greatly appreciate any help with this.

    Thanks,

    Jeff
     
  2. jcsd
  3. Jan 1, 2007 #2
    Step 1 is wrong.

    log ( A - B - C ) doesn't equal log ( A - B ) / log ( C )

    log ( A - B ) / log ( C ) is actually log [(A-B)/C]


    And I don't think the rest can be done.
     
  4. Jan 1, 2007 #3
    That is wrong too. The rule is log(A) - log(B) = Log(A/B).
     
  5. Jan 1, 2007 #4
    I made an error, but I still have the question...

    Okay, I made an error. I will step back to the original problem that I had but did not include.

    In...

    A - log ( e^(A-B) / e^(C)) = A - B - C

    What is C? If it cannot be done, could speculate or explain why it can't?

    Sorry about the confusion.

    Again, any help with this would be greatly appreciated.

    Thanks,

    Jeff
     
  6. Jan 1, 2007 #5
    Use the rules of logs, starting with the one I posted above, this should simplify very easily.
     
  7. Jan 1, 2007 #6

    cristo

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    Well, first subtract A from both sides. Next note that [tex]\frac{e^{(A-B)}}{e^{C}}=e^{(A-B)}e^{-C}=e^{(A-B-C)}[/tex] It should then simplify quite easily.
     
  8. Jan 1, 2007 #7
    Got it...

    Okay, thanks, guys.

    J
     
  9. Jan 1, 2007 #8
    and remember that log(e^x) = x
     
  10. Jan 3, 2007 #9

    Gib Z

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    Natural Log That is. I dont always like it when people use log for log base e, using ln is quicker and more informative...
     
  11. Jan 3, 2007 #10
    Yes, same here.
     
  12. Jan 4, 2007 #11

    uart

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    If by simplified he means to express C in closed form in terms of A and B then I'd like to see someone do it. It looks like a transendental equation to me.
     
  13. Jan 4, 2007 #12

    D H

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    It looks like a tautology to me.
     
  14. Jan 4, 2007 #13

    uart

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    tautology, how so?
     
  15. Jan 4, 2007 #14
    C=A/(ln10+1)-B

    correct me if I'm wrong.
     
    Last edited: Jan 4, 2007
  16. Jan 5, 2007 #15

    uart

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    You're wrong, just choose almost any two numbers for A and B and you'll find a counter-example. I tested your result with A=45 and B=3 and it definitely fails.
     
    Last edited: Jan 5, 2007
  17. Jan 5, 2007 #16

    HallsofIvy

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    Lambert's W function is defined as the inverse of the function f(x)= xex. It is, typically, the only way to solve an equation that has the unknown variable both as an exponent and not.
     
  18. Jan 5, 2007 #17

    I'm sorry but your example is correct.

    when A=45 and B=3 -> c=10.6256898

    In the original equation the left part is 31.3747102; the right part is 42-10.6256898 -> which is exactly 31.3747102.
     
  19. Jan 5, 2007 #18

    uart

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    No, the original equation was log( A - B - C ) = -C + A + B and those numbers definitely do not work in that equation.

    BTW, which equation are you substituting them into?
     
  20. Jan 5, 2007 #19
    Isn't that the original problem?

    That is the equation I solved..
     
  21. Jan 5, 2007 #20

    D H

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    The above simplifies to
    [tex]B+C = \log \left( \frac{e^{(A-B)}} {e^C}\right) = \log \left( {e^{(A-B-C)}}\right) = A-(B+C)[/tex]
    Solving for C,
    [tex]C=\frac A 2 - B[/tex]

    Your original post is
    This is an entirely different problem. So which statement of the problem is correct?
     
    Last edited: Jan 5, 2007
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