1) 1g (x-1)-1g (2x-11)=1g 2; 2) 1g (3x-1)-1g(x+5)=1g5; 3) log3 (x³-x) - log3 x = log3 3. 3391) ½ 1g (x²+x-5)=1g 5x+1g 1/5x; 2) ½ 1g (x²-4x-1)=1g 8x-lg 4x. 340 1) log3 (5x+3)=log3 (7x+5); 2) log₁/₂(3x-1)=log₁/₂(6x+8). 341 1) log₇ (x-1) log₇ x=log₇ x; 2) log₁/₃ x log₁/₃ (3x-2)=log₁/₃ (3x-2);

\[\lg (x-1) - \lg (2x-11) = \lg 2\]

\[\lg \frac{x-1}{2x-11} = \lg 2\]

\[\frac{x-1}{2x-11} = 2\]

\[x-1 = 4x - 22\]

\[3x = 21\]

\[x = 7\]

\[\lg (3x-1) - \lg (x+5) = \lg 5\]

\[\lg \frac{3x-1}{x+5} = \lg 5\]

\[\frac{3x-1}{x+5} = 5\]

\[3x-1 = 5x + 25\]

\[2x = -26\]

\[x = -13\]

\[\log_3 (x^3 - x) - \log_3 x = \log_3 3\]

\[\log_3 \frac{x^3 - x}{x} = \log_3 3\]

\[\frac{x^3 - x}{x} = 3\]

\[x^2 - 1 = 3\]

\[x^2 = 4\]

\[x = \pm 2\]

\[\frac{1}{2} \lg (x^2 + x - 5) = \lg 5x + \lg \frac{1}{5x}\]

\[\lg (x^2 + x - 5)^{\frac{1}{2}} = \lg (5x \cdot \frac{1}{5x})\]

\[\lg (x^2 + x - 5)^{\frac{1}{2}} = \lg 1\]

\[(x^2 + x - 5)^{\frac{1}{2}} = 1\]

\[x^2 + x - 5 = 1\]

\[x^2 + x - 6 = 0\]

\[x_1 = -3, x_2 = 2\]

\[\frac{1}{2} \lg (x^2 - 4x - 1) = \lg 8x - \lg 4x\]

\[\lg (x^2 - 4x - 1)^{\frac{1}{2}} = \lg \frac{8x}{4x}\]

\[\lg (x^2 - 4x - 1)^{\frac{1}{2}} = \lg 2\]

\[(x^2 - 4x - 1)^{\frac{1}{2}} = 2\]

\[x^2 - 4x - 1 = 4\]

\[x^2 - 4x - 5 = 0\]

\[x_1 = -1, x_2 = 5\]

\[\log_3 (5x+3) = \log_3 (7x+5)\]

\[5x + 3 = 7x + 5\]

\[2x = -2\]

\[x = -1\]

\[\log_{\frac{1}{2}} (3x-1) = \log_{\frac{1}{2}} (6x+8)\]

\[3x - 1 = 6x + 8\]

\[3x = -9\]

\[x = -3\]

\[\log_7 (x-1) \cdot \log_7 x = \log_7 x\]

\[\log_7 (x-1) = 1\]

\[x - 1 = 7\]

\[x = 8\]

\[\log_{\frac{1}{3}} x \cdot \log_{\frac{1}{3}} (3x-2) = \log_{\frac{1}{3}} (3x-2)\]

\[\log_{\frac{1}{3}} x = 1\]

\[x = \frac{1}{3}\]