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Thread: Logarithmic Equation

  1. #1
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    Logarithmic Equation

    Please help me solve these 2 logarithmic equations:



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  2. #2
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    Re: Logarithmic Equation

    $\ln\bigg[x^{3\ln{x}-2}\bigg] = \ln\bigg[\dfrac{x^{1+\ln{x}}}{e}\bigg]$

    $(3\ln{x}-2)\ln{x} = (1+\ln{x})\ln{x} - 1$

    $3\ln^2{x} - 2\ln{x} = \ln{x} + \ln^2{x} - 1$

    $2\ln^2{x} - 3\ln{x} + 1 = 0$

    $(2\ln{x} - 1)(\ln{x} - 1) = 0$

    can you finish?



    for the second, taking the base 2 log of each side yields ...

    $\dfrac{\log_2{x} - 6}{4} \cdot \log_2{x} = \log_2{4} - \log_2{x}$

    you'll get a quadratic in $\log_2{x}$ ...

    give it a go.
    Thanks from HallsofIvy and IloveIl
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  3. #3
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    Re: Logarithmic Equation

    Quote Originally Posted by IloveIl View Post
    Please help me solve these 2 logarithmic equations:
    1) OR rewrite as
    $ \begin{align*}e\cdot{\left( x \right)^{3\log (x) - 2}} &= {x^{1 + \log (x)}} \\ e\cdot{\left( x \right)^{2\log (x) - 3}} &= 1\\1 + \left( {{2{\log }^2}(x) - 3\log (x)} \right) &= 0\end{align*}$

    Now let $y=\log(x)$ we get $2y^2-3y+1=0$ to solve.

    Check Here.
    Last edited by Plato; Aug 25th 2017 at 04:44 PM.
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    Re: Logarithmic Equation

    for the second, taking the base 2 log of each side yields ...

    $\dfrac{\log_2{x} - 6}{4} \cdot \log_2{x} = \log_2{4} - \log_2{x}$

    you'll get a quadratic in $\log_2{x}$ ...

    give it a go.[/QUOTE]

    I didn't understand at all what you did here. can you show in more details?
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    Re: Logarithmic Equation

    Quote Originally Posted by IloveIl View Post
    for the second, taking the base 2 log of each side yields ...

    $\dfrac{\log_2{x} - 6}{4} \cdot \log_2{x} = \log_2{4} - \log_2{x}$

    you'll get a quadratic in $\log_2{x}$ ...

    give it a go.

    I didn't understand at all what you did here. can you show in more details?
    $\text {Let } u = log_2(x).$

    $x^{\frac{log_2(x) - 6}{4}} = \dfrac{4}{x} \implies log_2 \left ( x^{\frac{log_2(x) - 6}{4}} \right ) = log_2 \left ( \dfrac{4}{x} \right ) \implies$

    $ \dfrac{log_2(x) - 6}{4} * log_2(x) = log_2(4) - log_2(x) \implies \dfrac{u - 6}{4} * u = 2 - u \implies$

    $u^2 - 6u = 8 - 4u \implies what?$
    Last edited by JeffM; Aug 29th 2017 at 07:05 AM.
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  6. #6
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    Re: Logarithmic Equation

    Quote Originally Posted by IloveIl View Post
    for the second, taking the base 2 log of each side yields ...

    $\dfrac{\log_2{x} - 6}{4} \cdot \log_2{x} = \log_2{4} - \log_2{x}$

    you'll get a quadratic in $\log_2{x}$ ...

    give it a go.
    https://www.freemathhelp.com/forum/t..._2-x-6)-4)-4-x

    IloveIl, you never responded back at this site with any attempt to show that you understood those helpers.

    Had you followed up on that, then you might not be here asking the same questions.
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