Решить уравнение: 1. logx(2)−log4(x)+7/6=0 2.log3(3^x−8)=2-x

1. logx(2)−log4(x)+7/6=0, ОДЗ: x > 0 (log₂ 2 / log₂ x) - (1/2)*log₂ x + 7/6 = 0 1/(log₂ x) - (1/2)*log₂ x + 7/6 = 0 3log²₂ x - 7log₂ x - 6 = 0 Пусть log₂ x = z 3z² - 7z - 6 = 0 D = 49 + 4*3*6 = 121 z₁ = (7 - 11)/6 = - 1/3 z₂ = (7 + 11)/6 = 3 1) log₂ x = - 1/3 x = 2^(-1/3) x₁ = 1/∛2 2) log₂ x = 3 x₂ = 2³ x₂ = 8  2. log₃ (3^x−8 )= 2 - x, ОДЗ: 3^x - 8 > 0, 3^x > 8, x > log₃ 8 3^x - 8 = 3^(2 - x) 3^x - 8  = 9*(1/3^x) 3^(2x) - 8*(3^x) - 9 = 0 Пусть 3^x = z z² - 8z - 9 = 0 z₁ = -1 z₂ = 9 1)  3^x = - 1, не имеет смысла 2)  3^x = 9  3^x = 3² x = 2