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Math Help - Solving for x in a logarithm

  1. #1
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    (SOLVED) Solving for x in a logarithm

    log(base3)x + log(base3)(2x+51) = 4

    a) thank you for helping, I am a new member and love the looks of this site.
    b) i am not asking you to do this, maybe just some help I am stuck
    c) its not really homework, i am studying for a test on monday and just cant understand this

    1) I know the rules of logs and this is where i get stuck:

    log(base3)(2x+51)(x)=4
    log(base3)2x^2+51x=4
    2x^2+51x=3^4 --->2x^2+51x=81
    ? now do i divide both sides by 2? or should I have done that before i got rid of log(base3)?

    p.s. I know how to factor, but not something like 2x^2+51x-81=0

    Any help would be amazing, even a little input if your not sure to get me thinking. Thanks!
    Last edited by dpjwilson; March 20th 2009 at 08:13 PM. Reason: SOLVED
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  2. #2
    Junior Member
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    p.s. I know how to factor, but not something like 2x^2+51x-81=0
    2x^2 + 51x - 81 = 0

    (2x - 3)(x + 27)
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  3. #3
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    Wow, always getting caught up on the simple steps.
    Thanks for helping me out. When its something like that I always get thrown off a bit.
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  4. #4
    MHF Contributor
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    Quote Originally Posted by dpjwilson View Post
    2x^2+51x=3^4 --->2x^2+51x=81

    p.s. I know how to factor, but not something like 2x^2+51x-81=0
    Before your test, you might want to review some online lessons to refresh your memory regarding factoring quadratics.

    Good luck on the test!
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