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Math Help - Solve Logarithmic Equations

  1. #1
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    Solve Logarithmic Equations

    Needing help with a homework question. I know the answer but do not know how to show my work to get that answer:
    Ln(4x^2-3x)=ln(16x-12)-ln x
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  2. #2
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    Re: Solve Logarithmic Equations

            Ln(4x^2-3x)=ln(16x-12)-ln x This can be written as :         Ln(4x^2-3x)=ln ( \frac{(16x-12)}{x}) { using property of log a - log b = log (a/b) } Now we can compare both sides we get : 4x^2-3x =  \frac{(16x-12)}{x} Now you can easily solve this equation. Hope you understood
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  3. #3
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    Re: Solve Logarithmic Equations

    Hello, paigesisco!

    \text{Solve: }\:\ln(4x^2-3x)\:=\:\ln(16x-12)-\ln x

    Note that: . 16x-12\:>\:0 \quad\Rightarrow\quad x \:>\:\tfrac{3}{4}


    We have: . \ln(4x^2-3x) \ln x \:=\:\ln(16x-12)

    . . . . . . . . \ln\left[x(4x^2 - 3x)\right] \:=\:\ln(16x-12)

    . . . . . . . . . . \ln(4x^3 - 3x^2) \:=\:\ln(16x-12)

    . . . . . . . . . . . . 4x^3 - 3x^2 \:=\:16x-12

    . . . . 4x^3 - 3x^2 - 16x + 12 \:=\:0

    . . . x^2(4x-3) - 4(4x - 3) \:=\:0

    . . . . . . . (x^2-4))(4x-3) \:=\:0

    Hence: . x \;=\;\pm2,\,\tfrac{3}{4}


    The only root in the domain is: . x = 2
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