1. ## logarithamic eqns

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

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

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

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

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

Cancel the logs and solve for x ..

3. 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. 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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