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Logarithm equation

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    log3(3x-5) + log(1/3) 12 = 1 - x


    2. Relevant equations
    logarithm
    exponential


    3. The attempt at a solution
    The best I can get is:

    [tex]\frac{3x-5}{12}=3^{1-x}[/tex]

    Can this be solved??

    Thanks
     
  2. jcsd
  3. Nov 8, 2012 #2

    symbolipoint

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    You have two different bases and what you did to get your equation in part three ignored the two different bases. Use the change-of-base formula.

    [itex]\[
    \log _b x = \frac{{\log _k x}}{{\log _k b}}
    \]
    [/itex]

    [itex]\[
    \log _{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 3$}}} 12 = \frac{{\log _3 12}}{{\log _3 \left( {{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 3$}}} \right)}}
    \]
    [/itex]

    Please excuse the slightly bad formatting result in lefthand member of last equation. "Logarithm to the base one-third of twelve...."
     
    Last edited: Nov 8, 2012
  4. Nov 8, 2012 #3
    This is what I did:

    log3(3x-5) + log(1/3) 12 = 1 - x

    log3(3x-5) + log(3-1)12 = 1 - x

    log3(3x-5) - log312 = 1 - x

    log3 [(3x-5)/12] = 1 - x

    and the last line is the same as what I post in #1


    If I understand it correctly, I think I will come up with the same equation as mine using your idea. Am I correct?

    Thanks
     
  5. Nov 9, 2012 #4

    ehild

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    That can be solved only numerically. Are you sure, it was not log3(3x-5) + log(1/3) 12 = 1 - x?

    ehild
     
  6. Nov 9, 2012 #5

    symbolipoint

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    Your second line is wrong. Look again at the change of base formula.
     
  7. Nov 9, 2012 #6

    symbolipoint

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    songoku did not say if he is trying to solve the equation or not. If so, you're correct, but he may still want a transformed version of the original logarithm equation.

    ... In fact, the transformed equation in exponential form can be simplified significantly, still having an x in an exponent and and x term in a polynomial.
     
    Last edited: Nov 9, 2012
  8. Nov 9, 2012 #7

    ehild

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    Songoku did the transformation correctly. What is log3(1/3) ?

    ehild
     
  9. Nov 9, 2012 #8

    symbolipoint

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    ?
    No.
     
  10. Nov 9, 2012 #9
    I am sure. I have checked it again.

    Why is it wrong?

    1/3 = 3-1

    and if I have log(an)b I can change it to 1/n . loga b, correct?
     
  11. Nov 9, 2012 #10

    ehild

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    It can be a typo in the book then.
    You can find approximate solution graphically.


    ehild
     
  12. Nov 9, 2012 #11

    ehild

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    Well, how do you transform the equation then? Show your result.

    ehild
     
  13. Nov 9, 2012 #12

    symbolipoint

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    This is shown in post #2 for the treatment of the term that used the rational base of one-third.
     
  14. Nov 9, 2012 #13
    OK let me test it

    log3(3x-5) + log(1/3) 12 = 1 - x

    Use change of base rule, this becomes:

    log3(3x-5) + [log3 12] / [log3 (1/3)]= 1 - x

    log3(3x-5) + [log3 12] / (-1)= 1 - x

    log3(3x-5) - [log3 12] = 1 - x ---> same as what I got ???
     
  15. Nov 9, 2012 #14

    symbolipoint

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    songoku, okay you convinced me. Although you did not quite finish, the work looks good.
     
  16. Nov 9, 2012 #15
    ehild is correct, there can be a typo in the book. Plug the equation in wolframalpha.
     
  17. Nov 9, 2012 #16

    SammyS

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    Using that treatment you also get songoku's result of
    [itex]\displaystyle \frac{3x-5}{12}=3^{1-x}[/itex]​
    which is correct, as ehild has pointed out more than once.
     
  18. Nov 9, 2012 #17

    symbolipoint

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    I made a small but very significant arithmetic mistake when I made the equation transformation myself the first time. The equation result which you show is good.
     
  19. Nov 9, 2012 #18
    Thanks for all the help given here :smile:
     
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