3. Решить уравнение: 1) log2(2+log3(3+x)) = 0 [lg(3+2log2(1+x)) = 0]; 2) 3log3x+3logx3=10 [3log7x-2logx7 = 1].

Первый случай: 1) log2(2+log3(3+x)) = 0 2 + log3(3+x) = 2^0 = 1 log3(3+x) = 1 - 2 = -1 3 + x = 3^{-1} = \frac{1}{3} x = \frac{1}{3} - 3 = \frac{1 - 9}{3} = -\frac{8}{3} Проверим: 3 + x = 3 - \frac{8}{3} = \frac{9 - 8}{3} = \frac{1}{3} > 0 log_3(3+x) = log_3(\frac{1}{3}) = -1 2 + log_3(3+x) = 2 - 1 = 1 > 0 Ответ: x = -8/3 Второй случай: [lg(3+2log2(1+x)) = 0] 3 + 2log2(1+x) = 10^0 = 1 2log2(1+x) = 1 - 3 = -2 log2(1+x) = -1 1 + x = 2^{-1} = \frac{1}{2} x = \frac{1}{2} - 1 = -\frac{1}{2} Проверим: 1 + x = 1 - \frac{1}{2} = \frac{1}{2} > 0 Ответ: x = -1/2 Третий случай: 2) 3log3x+3logx3=10 3log_3 x + \frac{3}{log_3 x} = 10 Пусть y = log_3 x 3y + \frac{3}{y} = 10 3y^2 + 3 = 10y 3y^2 - 10y + 3 = 0 D = (-10)^2 - 4*3*3 = 100 - 36 = 64 y1 = (10 + \sqrt{64}) / (2 * 3) = (10 + 8) / 6 = 18/6 = 3 y2 = (10 - \sqrt{64}) / (2 * 3) = (10 - 8) / 6 = 2/6 = 1/3 log_3 x = 3 -> x = 3^3 = 27 log_3 x = 1/3 -> x = 3^(1/3) = \sqrt[3]{3} Ответ: x = 27, x = \sqrt[3]{3} Четвертый случай: [3log7x-2logx7 = 1] 3log_7 x - \frac{2}{log_7 x} = 1 Пусть y = log_7 x 3y - \frac{2}{y} = 1 3y^2 - 2 = y 3y^2 - y - 2 = 0 D = (-1)^2 - 4 * 3 * (-2) = 1 + 24 = 25 y1 = (1 + \sqrt{25}) / (2 * 3) = (1 + 5) / 6 = 6/6 = 1 y2 = (1 - \sqrt{25}) / (2 * 3) = (1 - 5) / 6 = -4/6 = -2/3 log_7 x = 1 -> x = 7^1 = 7 log_7 x = -2/3 -> x = 7^(-2/3) = \frac{1}{\sqrt[3]{7^2}} = \frac{1}{\sqrt[3]{49}} Ответ: x = 7, x = \frac{1}{\sqrt[3]{49}}