1. Solve the equation $5) \log_2 x + \log_2 (x - 1) = 1$. - Use the logarithm property: $\log_b a + \log_b c = \log_b (ac)$. - So, $\log_2 [x(x - 1)] = 1$. - Rewrite in exponential form: $x(x - 1) = 2^1 = 2$. - Expand: $x^2 - x = 2$. - Rearrange: $x^2 - x - 2 = 0$. - Factor: $(x - 2)(x + 1) = 0$. - Solutions: $x = 2$ or $x = -1$. - Check domain: $x > 0$ and $x - 1 > 0 \Rightarrow x > 1$. - So, valid solution is $x = 2$. 2. Solve the equation $6) 36^x - 6^x - 6 = 0$. - Express $36$ as $6^2$: $36^x = (6^2)^x = 6^{2x}$. - Let $y = 6^x$. - Rewrite: $y^2 - y - 6 = 0$. - Factor: $(y - 3)(y + 2) = 0$. - Solutions: $y = 3$ or $y = -2$. - Since $y = 6^x > 0$, discard $y = -2$. - Solve $6^x = 3$. - Take $ log_6$ both sides: $x = \log_6 3$. 3. Solve for $x$ in $\log x = 7 \log x + \log y - 2 \log z$. - Rearrange: $\log x - 7 \log x = \log y - 2 \log z$. - Simplify left: $-6 \log x = \log y - 2 \log z$. - Rewrite right side using log properties: $\log y - \log z^2 = \log \frac{y}{z^2}$. - So, $-6 \log x = \log \frac{y}{z^2}$. - Multiply both sides by $-1$: $6 \log x = -\log \frac{y}{z^2} = \log \left(\frac{y}{z^2}\right)^{-1} = \log \frac{z^2}{y}$. - Divide both sides by 6: $\log x = \frac{1}{6} \log \frac{z^2}{y}$. - Use power rule: $\log x = \log \left(\frac{z^2}{y}\right)^{\frac{1}{6}}$. - Therefore, $x = \left(\frac{z^2}{y}\right)^{\frac{1}{6}}$. 4. Logarithm of the expression $\log \sqrt[9]{\frac{ab}{c^2}}$. - Rewrite root as power: $\sqrt[9]{\frac{ab}{c^2}} = \left(\frac{ab}{c^2}\right)^{\frac{1}{9}}$. - Apply log power rule: $\log \left(\frac{ab}{c^2}\right)^{\frac{1}{9}} = \frac{1}{9} \log \frac{ab}{c^2}$. - Use log quotient rule: $\frac{1}{9} (\log ab - \log c^2)$. - Use log product and power rules: $\frac{1}{9} (\log a + \log b - 2 \log c)$. Final answers: $5)\ x=2$ $6)\ x=\log_6 3$ $3)\ x=\left(\frac{z^2}{y}\right)^{\frac{1}{6}}$ $4)\ \log \sqrt[9]{\frac{ab}{c^2}} = \frac{1}{9} (\log a + \log b - 2 \log c)$