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Solve for x log3 (x-2) + log3 (x-4) = 7

  1. Jul 6, 2011 #1
    1. The problem statement, all variables and given/known data
    i have been trying to do this ques. 2 days now and still cant get the ans.
    help plz

    Solve for x
    log3 (x-2) + log3 (x-4) = 7




    2. Relevant equations



    3. The attempt at a solution
    log3 (x-2) (x-4) = 7
     
  2. jcsd
  3. Jul 6, 2011 #2

    berkeman

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    Re: logarithms

    A reply from another thread:

     
  4. Jul 6, 2011 #3
    Re: logarithms

    what does that mean
     
  5. Jul 6, 2011 #4

    berkeman

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    Re: logarithms

    His hint was to use the inverse function of the logarithm as your first step.

    So if you were working with log base 2, you would raise both sides to the exponent power of 2....

    [tex]log_2 (x) = 5[/tex]

    [tex]2^{log_2 (x)} = 2^5[/tex]

    ...and simplify the lefthand side...
     
  6. Jul 6, 2011 #5
    Re: logarithms

    i would then prolly do this

    x^2-4x-2x-8 = 3^7

    x^2- 6x- 8 = 3^7

    am i on the right track
     
  7. Jul 6, 2011 #6

    berkeman

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    Re: logarithms

    There's a sign error in your multiplication of (-2)*(-4), but other than that...
     
  8. Jul 6, 2011 #7
    Re: logarithms

    iam lose, dont know where to go frm here
     
  9. Jul 6, 2011 #8

    berkeman

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  10. Jul 6, 2011 #9
    Re: logarithms

    x^2 - 6x + 8 = 3^7

    x^2 - 6x + 8= 2187

    is that good
     
  11. Jul 6, 2011 #10

    SammyS

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    Re: logarithms

    Looks good. Now solve for x.
     
  12. Jul 6, 2011 #11

    berkeman

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    Re: logarithms

    And once you solve for x, use your scientific calculator to check your answer by plugging it back into your original problem statement in Post #1.
     
  13. Jul 6, 2011 #12
    Re: logarithms

    x^2 - 6x + 8= 2187

    x^2 - 6x +8 - 2187 =0

    x^2 - 6x - 2179 =0

    x = - 6 [itex]\pm[/itex] [itex]\sqrt{}[/itex] 6^2 - 4*1*2179[itex]/[/itex] 2*1

    x = - 6 ± [itex]\sqrt{}[/itex] 36 - 8716 [itex]/[/itex] 2*1

    x = - 6 ± [itex]\sqrt{}[/itex] -8680 [itex]/[/itex] 2

    x = -6 ± 93.167[itex]/[/itex] 2

    x = -6 + 93.167[itex]/[/itex] 2

    x = 87.167 [itex]/[/itex] 2

    X = 4.436

    how is that so far
     
  14. Jul 6, 2011 #13

    berkeman

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    Re: logarithms

    You tell us. Is the answer correct when you plug it into the original equation in Post #1?


    EDIT -- I do think I see a sign error in the first line where you apply the Quadratic Formula, though...
     
  15. Jul 6, 2011 #14
    Re: logarithms

    i dont know how to plug it into the equation
     
  16. Jul 6, 2011 #15

    berkeman

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    Re: logarithms

    First, fix the sign error(s) in the Quadratic Equation application, and then, do you have a scientific calculator that lets you do log base n?
     
  17. Jul 6, 2011 #16

    berkeman

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  18. Jul 6, 2011 #17
    Re: logarithms

    wen one that i am using is log 10
     
  19. Jul 6, 2011 #18
    Re: logarithms

    that link does not work and what error is in the eqaution
     
  20. Jul 6, 2011 #19
    Re: logarithms

    how do i substitue it
     
  21. Jul 6, 2011 #20

    berkeman

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    Re: logarithms

    The link works fine when I click on it. But in any case, all you have to do is Google logarithm calculator, and you get lots of useful links (inclucding that one).

    You have two sign errors that I think I see in your application of the Quadratic Equation. Just be careful in handling negative signs, and re-check the line where you substitute A, B and C into the Quadratic Equation...
     
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