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Math Help - solve logarithmic equation

  1. #1
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    solve logarithmic equation

    ln(x-2)+ln(2x-3)=2lnx

    solve for x. My brain must have gone to sleep, i have the answer but can't understand why i can't use:

    e^ln(x-2) + e^ln(2x-3) = e^lnx^2

    (x-2) + (2x-3) = x^^2

    and then solve quadratic
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  2. #2
    ynj
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    Quote Originally Posted by attwoodp View Post
    ln(x-2)+ln(2x-3)=2lnx

    solve for x. My brain must have gone to sleep, i have the answer but can't understand why i can't use:

    e^ln(x-2) + e^ln(2x-3) = e^lnx^2

    (x-2) + (2x-3) = x^^2

    and then solve quadratic
    e^ln(x-2) + e^ln(2x-3) = e^lnx^2

    (x-2) + (2x-3) = x^^2
    it should be
    e^ln(x-2) * e^ln(2x-3) = e^lnx^2

    (x-2)* (2x-3) = x^^2
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  3. #3
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    Quote Originally Posted by attwoodp View Post
    ln(x-2)+ln(2x-3)=2lnx

    solve for x. My brain must have gone to sleep, i have the answer but can't understand why i can't use:

    e^ln(x-2) + e^ln(2x-3) = e^lnx^2

    (x-2) + (2x-3) = x^^2

    and then solve quadratic
    You can't do that; what you meant to do is raise both sides to be the exponent of e, and then:

    e^{ln(x-2)+ln(2x-3)} = e^{2lnx} \Rightarrow e^{ln( (x-2)(2x-3) )} = e^{ln (x^2)}

    This gives (x-2)(2x-3) = x^2 , and you know the rest...
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  4. #4
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    Quote Originally Posted by attwoodp View Post
    ln(x-2)+ln(2x-3)=2lnx solve for x.
    Use log rules to convert the equation to:

    \ln(2x^2\, -\, 7x\, +\, 6)\, =\, \ln(x^2)

    Then equate the arguments, and solve the resulting quadratic equation.
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