10) logs²x-log5 x = 2

10) log₅²x - log₅x = 2 Пусть log₅x = t, тогда t² - t = 2 t² - t - 2 = 0 D = 1 + 8 = 9 t₁ = (1 + 3)/2 = 2 t₂ = (1 - 3)/2 = -1 Обратная замена: log₅x = 2 => x = 5² = 25 log₅x = -1 => x = 5⁻¹ = 1/5 = 0.2 Ответ: 0.2; 25