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Solve for a variable in the exponent

  1. Jan 21, 2009 #1
    I can't remember how to solve an equation of the following form:

    [tex]ln(a^x) = ln(b^x)[/tex]

    I want to solve this for x where a and b are both real numbers. I know I can manipulate the logarithm to write:

    [tex]xln(a) = xln(b)[/tex]

    But this just seems to cancel out the x's. If you can show me how to solve this type of equation for x, I would very much appreciate it (it's not homework, just trying to remember how to do it).

    Thank you for your time.
  2. jcsd
  3. Jan 21, 2009 #2
    As it should, ln(a^x) = ln(b^x) iff a = b
  4. Jan 21, 2009 #3
    Good point. I think what I meant to say was:

    [tex]C*ln(a^x) = D*ln(b^x)[/tex] where a,b,C,D are all real numbers.

    This is currently what I'm thinking about it...

    [tex]xC*ln(a) = xD*ln(b)[/tex]
    [tex]xC*ln(a) - xD*ln(b) = 0[/tex]
    [tex]x*(C*ln(a) - D*ln(b)) = 0[/tex]

    It seems close, but I can't quite find x.
    Last edited: Jan 21, 2009
  5. Jan 21, 2009 #4


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    Just factorize x out. If you ever feel like you want to divide through by the variable - stop - and rather factorize x out :wink:
    This means x would be equal to 0 and possibly other values too. When you divide through by x, the denominator can never equal 0 therefore you lose your x=0 answer.

    [tex]Cx ln a - Dx ln b=0[/tex]

    [tex]x(C ln a - D ln b)=0[/tex]
  6. Jan 21, 2009 #5


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    I think you have one too many f's in that statement.:wink: It is also true for all 'a' and 'b' if x=0.
  7. Jan 21, 2009 #6
    It still looks like x can be anything though. Maybe I'm still phrasing the problem wrong. I remember doing something like this a long time ago and x had a definite value...

    Thanks for your help anyway.
  8. Jan 21, 2009 #7


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    It would look like (and be) true for all x if:

    [tex]ln(a^x)=ln(b^x)[/tex] where [tex]a=b[/tex]

    But the only true value for x for any real value of a and b is 0.
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