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2log(x-1) + logx = logx + log4

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data

    2log(x-1) + logx = logx + log4


    2. Relevant equations



    3. The attempt at a solution
    log(x-1)^2 + logx = logx (4)
     
  2. jcsd
  3. Jul 7, 2011 #2
    Re: logs

    Hey there,

    ok it looks like you have some exact same term on both sides to begin with, so those can just cancel out right? This leaves you with a log of something = a log of something else, that should simplify it down nice for ya.
     
  4. Jul 7, 2011 #3
    Re: logs

    i dont get it
     
  5. Jul 7, 2011 #4
    Re: logs

    so are u saying it should be:
    log (2x - x)^2 = 4x
     
  6. Jul 7, 2011 #5
    Re: logs

    Well the first step is just an algebra step, you have log (x) on both sides of the = sign, so the equation can be reduced. When that happens you are left with a log (of something) = log (of something else) when this happens you can sort of drop the log and just solve for x.
     
  7. Jul 7, 2011 #6
    Re: logs

    Noooooo, :)
    What you need to do here is just first cancel out the terms [itex]\log x[/itex] on both sides.
    Then shift [itex]\log 4[/itex] to the left side remaining 0 at the right side.
    Now using these two identities, try to figure out the answer:-
    [tex]p(\log_a b)=log_a b^p[/tex]
    and
    [tex]\log_a u-\log_a v=\log_a \frac{u}{v}[/tex]
     
  8. Jul 7, 2011 #7
    Re: logs

    log (x-1) = log4

    i dont think i understand that
     
  9. Jul 7, 2011 #8
    Re: logs

    2log (x-1) - log 4 = 0
     
  10. Jul 7, 2011 #9
    Re: logs

    There you go! I'd have kept log(4) on the right side and then just solve for x.
     
  11. Jul 7, 2011 #10
    Re: logs

    Yep, that's what i meant to say. Now try converting 2log (x-1) using the identities i gave. :smile:
     
  12. Jul 7, 2011 #11
    Re: logs

    log (x-1)^2 - log 4 = 0
     
  13. Jul 7, 2011 #12
    Re: logs

    Now use the second identity. :wink:
     
  14. Jul 7, 2011 #13
    Re: logs

    log (x^2-1) /4 = 0
     
  15. Jul 7, 2011 #14
    Re: logs

    It would be
    [tex]\log \frac{(x-1)^2}{4}=0[/tex]

    Whatever the base would be, when you use the antilog method here, the right hand side would become 1. So what you get is:-
    [tex]\frac {(x-1)^2}{4}=1[/tex]

    Now solve it further. :wink:
     
  16. Jul 7, 2011 #15
    Re: logs

    ok so i would do this next but i dont think its right

    (x^2-1) /4 = 1
     
  17. Jul 7, 2011 #16
    Re: logs

    I don't understand how you convert (x-1)2 to (x2-1)? :confused:
    That's wrong!!

    Solving this equation:-
    [tex]\frac {(x-1)^2}{4}=1[/tex]
    we get,
    [tex](x-1)^2=4[/tex]
    Now try to do it.
     
  18. Jul 7, 2011 #17
    Re: logs

    ok, but i thought (x-1) meant (x-1) (x-1)
     
  19. Jul 7, 2011 #18
    Re: logs

    I still didn't get you :confused:
    [tex](x-1)^2=(x-1)(x-1)=x^2-2x+1[/tex]
     
  20. Jul 7, 2011 #19
    Re: logs

    so i would end up with:

    x^2-2x+1 = 4
     
  21. Jul 7, 2011 #20
    Re: logs

    and then i would have to do the quadratic equation, right
     
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