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Logarithm equation

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Express 2log[tex]_{2}X[/tex]=1+log[tex]_{a}(7X-10a)[/tex] find x in terms of a.
    i wondering if there is other methods to solve aside from completing the sq.


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations
    I got x²-7ax+10a²=0
    (x-[tex]\frac{7a}{2}[/tex])²-10a²-([tex]\frac{7a}{2}[/tex])²=0
    x-[tex]\frac{7a}{2}[/tex]=±[tex]\frac{3a}{2}[/tex]
    x=5a or x=2a
     
    Last edited: Nov 23, 2008
  2. jcsd
  3. Nov 23, 2008 #2
    Re: Logarithm

    Start by:
    2logXX/logX2=1+logX(7X-10a)/logXa
    After doing some transformations, I came up with:
    2/logX2=logX((a(7x-10a))/logXa
    Out of here:
    logX((a(7x-10a))=logXX2
    and
    logX2=logXa
    So we got 7ax-10a2=x2 and a=2
    I think now it is easy to go on.
     
  4. Nov 23, 2008 #3
    Re: Logarithm

    your step is wrong.. could you double chk
     
  5. Nov 23, 2008 #4
    Re: Logarithm

    Could you possibly tell me what step is wrong?
     
  6. Nov 23, 2008 #5
    Re: Logarithm

    Should not be
    2/logX2=logX((a(7x-10a))/logXa
    instead
    2/logX2=logX(x(7x-10a)/logXa
     
  7. Nov 23, 2008 #6
    Re: Logarithm

    If you think about this step, it is correct:
    1+logX(7X-10a)/logXa
    logXa+logX(7X-10a)/logXa
    out of there:
    logX(a*(7X-10a)/logXa
     
    Last edited: Nov 23, 2008
  8. Nov 23, 2008 #7
    Re: Logarithm

    how can logXa be 1 shud be logxX mahs
     
  9. Nov 23, 2008 #8
    Re: Logarithm

    [tex]1+\frac{log_x(7X-10a)}{log_xa}[/tex]

    [tex]\frac{log_xa+log_x(7X-10a)}{log_xa}[/tex]

    Do you understand, now?

    Regards.
     
  10. Nov 23, 2008 #9
    Re: Logarithm

    i noe wat you mean..

    only logxX can be 1
    logxA cannot be 1 as x is not equal to A
     
  11. Nov 23, 2008 #10
    Re: Logarithm

    You don't know what I mean.

    I never said that logxA=1

    Try solving the whole equation using my method. I already gave you pretty much information.

    You couldn't understand that logxa/logxa=1 ?

    The end.
     
  12. Nov 23, 2008 #11
    Re: Logarithm

    if tats the case

    logX((a(7x-10a))=logXX2
    is still wrong bah
    cos shud be 2log2( A ) instead of logXX2
     
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