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Thread: logarithamic eqns

  1. #1
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    logarithamic eqns

    2lg(x+2)-lg x =4lg3 -lg4 [solve for x]


    Log3[3^(x+1) 18 sq root 3] =x [solve for x]

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  2. #2
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    $\displaystyle 2lg(x+2)-lg x =4lg3 -lg4$

    $\displaystyle lg(x+2)^2-lg x=lg3^4-lg4$

    $\displaystyle
    lg(\frac{(x+2)^2}{x})=lg(\frac{3^4}{4})
    $

    Cancel the logs and solve for x ..

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  3. #3
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    Do you mean this?

    $\displaystyle 2\log(x+2) - \log x = 4\log 3 - \log 4$

    If so...

    $\displaystyle \begin{aligned}
    2\log(x+2) - \log x &= 4\log 3 - \log 4 \\
    \log (x+2)^2 - \log x &= \log 3^4 - \log 4 \\
    \log \left(\frac{(x+2)^2}{x}\right) &= \log \left(\frac{81}{4}\right) \\
    \frac{(x+2)^2}{x} &= \frac{81}{4}
    \end{aligned}$

    $\displaystyle \begin{aligned}
    4(x + 2)^2 &= 81x \\
    4(x^2 + 4x + 4) &= 81x \\
    4x^2 + 16x + 16 &= 81x \\
    4x^2 - 65x + 16 &= 0 \\
    (4x - 1)(x - 16) &= 0
    \end{aligned}$

    x = 1/4 or x = 16


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  4. #4
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    Again, do you mean this?

    $\displaystyle \log_3[3^{x+1} - 18\sqrt{3}] = x$

    If so...

    $\displaystyle \begin{aligned}
    \log_3[3^{x+1} - 18\sqrt{3}] &= x \\
    3^x &= 3^{x+1} - 18\sqrt{3} \\
    18\sqrt{3} &= 3^{x+1} - 3^x \\
    2(3^2)(3^{1/2}) &= 3(3^x) - 3^x
    \end{aligned}$

    $\displaystyle \begin{aligned}
    2(3^{5/2}) &= 3^x (3 - 1) \\
    2(3^{5/2}) &= 2(3^x) \\
    3^{5/2} &= 3^x \\
    x = \frac{5}{2}
    \end{aligned}$


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