6. Решить уравнение (\sqrt[3]{x})log7x-2=7 [(√x)lgx = 104+lgx].

Первый случай: (\sqrt[3]{x})^{log_7 x - 2} = 7 x^{\frac{1}{3}(log_7 x - 2)} = 7 log_7 (x^{\frac{1}{3}(log_7 x - 2)}) = log_7 7 \frac{1}{3}(log_7 x - 2) log_7 x = 1 (log_7 x - 2) log_7 x = 3 (log_7 x)^2 - 2 log_7 x - 3 = 0 Пусть y = log_7 x y^2 - 2y - 3 = 0 D = (-2)^2 - 4 * 1 * (-3) = 4 + 12 = 16 y1 = (2 + \sqrt{16}) / (2 * 1) = (2 + 4) / 2 = 6 / 2 = 3 y2 = (2 - \sqrt{16}) / (2 * 1) = (2 - 4) / 2 = -2 / 2 = -1 log_7 x = 3 -> x = 7^3 = 343 log_7 x = -1 -> x = 7^{-1} = \frac{1}{7} Ответ: x = 343, x = 1/7 Второй случай: [(√x)lgx = 104+lgx]. (\sqrt{x})^{lg x} = 10^{4 + lg x} (x^{\frac{1}{2}})^{lg x} = 10^{4 + lg x} x^{\frac{1}{2}lg x} = 10^{4 + lg x} lg (x^{\frac{1}{2}lg x}) = lg (10^{4 + lg x}) \frac{1}{2} lg x * lg x = (4 + lg x) * lg 10 \frac{1}{2} (lg x)^2 = 4 + lg x Пусть y = lg x \frac{1}{2} y^2 = 4 + y y^2 = 8 + 2y y^2 - 2y - 8 = 0 D = (-2)^2 - 4 * 1 * (-8) = 4 + 32 = 36 y1 = (2 + \sqrt{36}) / (2 * 1) = (2 + 6) / 2 = 8 / 2 = 4 y2 = (2 - \sqrt{36}) / (2 * 1) = (2 - 6) / 2 = -4 / 2 = -2 lg x = 4 -> x = 10^4 = 10000 lg x = -2 -> x = 10^{-2} = \frac{1}{100} = 0.01 Ответ: x = 10000, x = 0.01