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Math Help - 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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     2<b>lg</b>(x+2)-<b>lg</b> x =4<b>lg</b>3 -<b>lg</b>4

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

     <br />
lg(\frac{(x+2)^2}{x})=lg(\frac{3^4}{4})<br />

    Cancel the logs and solve for x ..

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  3. #3
    Super Member
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    Do you mean this?

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

    If so...

    \begin{aligned}<br />
2\log(x+2) - \log x &= 4\log 3 - \log 4 \\<br />
\log (x+2)^2 - \log x &= \log 3^4 - \log 4 \\<br />
\log \left(\frac{(x+2)^2}{x}\right) &= \log \left(\frac{81}{4}\right) \\<br />
\frac{(x+2)^2}{x} &= \frac{81}{4}<br />
\end{aligned}

    \begin{aligned}<br />
4(x + 2)^2 &= 81x \\<br />
4(x^2 + 4x + 4) &= 81x \\<br />
4x^2 + 16x + 16 &= 81x \\<br />
4x^2 - 65x + 16 &= 0 \\<br />
(4x - 1)(x - 16) &= 0<br />
\end{aligned}

    x = 1/4 or x = 16


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  4. #4
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    Again, do you mean this?

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

    If so...

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

    \begin{aligned}<br />
2(3^{5/2}) &= 3^x (3 - 1) \\<br />
2(3^{5/2}) &= 2(3^x) \\<br />
3^{5/2} &= 3^x \\<br />
x = \frac{5}{2}<br />
\end{aligned}


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