1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Mentallic

    User Avatar
    Homework Helper

    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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    Mentallic

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Solve for a variable in the exponent
Loading...