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Thread: log

  1. #1
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    Lightbulb log

    18. Solve the equation e^(2x) = 3

    19. Solve the equation ln (3x - 2) + ln (x - 1) = 2 ln x.
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  2. #2
    Junior Member Pinsky's Avatar
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    $\displaystyle e^{2x}=3$
    then you ln the whole equation. And get
    $\displaystyle x=\frac{1}{2}\ln{3}$


    $\displaystyle ln{(3x-2)}+ln{(x-1)}=2\ln{x}$
    $\displaystyle ln{(3x-2)(x-1)}=\ln{x^2}$
    $\displaystyle (3x-2)(x-1)=x^2$
    $\displaystyle 2x^2-5x+2=0$
    $\displaystyle x_1=2$
    $\displaystyle x_2=\frac{1}{2}$

    you solve the quadratic equation and get results.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Pinsky View Post
    $\displaystyle ln{(3x-2)}+ln{(x-1)}=2\ln{x}$
    $\displaystyle ln{(3x-2)(x-1)}=\ln{x^2}$
    $\displaystyle (3x-2)(x-1)=x^2$
    $\displaystyle 2x^2-5x+2=0$
    $\displaystyle x_1=2$
    $\displaystyle x_2=\frac{1}{2}$

    you solve the quadratic equation and get results.
    That's a nice derivation. But did you notice that $\displaystyle x_2 = \frac{1}{2}$ is not a solution? It doesn't lie in the domain of the original expression.

    -Dan
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  4. #4
    Junior Member Pinsky's Avatar
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    See it now, tnx.
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