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    i honestly thought I had this sorted, but it seems not.

    the question =
    log(base 18) x + log(base x) 3 = 2

    my method:

    log(base 18) x becomes logx / log18
    log(base x) 3 becomes log3/logx



    as its addition, I multiplied the fractioned versions and took out log

    log [ ( x/18 ) / (3/x) ] = 2

    But the x's cancel, and I need them to solve.
    Hence, does it mean there is no solution? :/
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    The question is unclear to me, sorry...
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    (Original post by TheSadReaper)
    The question is unclear to me, sorry...
    sorry.
    find the value (s) of x for which:

    log_18  x + log_x 3 = 2

    its not log (base 1) 8x...i cant get latex to show it as base 18
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    log_{18}x+log_x3=2

    log_{18}x+\dfrac{log_{18}3}{log_  {18}x}=2

    (log_{18}x)^2+log_{18}3=2log_{18  }x
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    (Original post by Pheylan)
    log_{18}x+log_x3=2

    log_{18}x+\dfrac{log_{18}3}{log_  {18}x}=2

    (log_{18}x)^2+log_{18}3=2log_{18  }x
    Who's that in your sig? Dayumm!
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    \log_a x = \frac{1}{ \log_x a}

    So  \frac {1}{ \log_x 18} + \log_x 3 = 2

    I think that's right
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    (Original post by Pheylan)
    log_{18}x+log_x3=2

    log_{18}x+\dfrac{log_{18}3}{log_  {18}x}=2

    (log_{18}x)^2+log_{18}3=2log_{18  }x
    EDITING
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    (Original post by Pheylan)
    log_{18}x+log_x3=2

    log_{18}x+\dfrac{log_{18}3}{log_  {18}x}=2

    (log_{18}x)^2+log_{18}3=2log_{18  }x
    The LHS becomes
     log_18 (x^2)
    i'm not sure about the RHS but I do think I can take out a factor of  log_{18}
    help appreciated!
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    (Original post by dream123)
    The LHS becomes
     log_18 (x^2)
    i'm not sure about the RHS but I do think I can take out a factor of  log_{18}
    help appreciated!
    no

    (log_{18}x)^2+log_{18}3=2log_{18  }x

    (log_{18}x)^2-2log_{18}x+log_{18}3=0

    what sort of equation is this?
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    (Original post by Pheylan)
    no

    (log_{18}x)^2+log_{18}3=2log_{18  }x

    (log_{18}x)^2-2log_{18}x+log_{18}3=0

    what sort of equation is this?
    Quadratic, but it won't solve :/
 
 
 
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