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 = 01/(log₂ x) - (1/2)*log₂ x + 7/6 = 03log²₂ x - 7log₂ x - 6 = 0Пусть log₂ x = z3z² - 7z - 6 = 0D = 49 + 4*3*6 = 121z₁ = (7 - 11)/6 = - 1/3z₂ = (7 + 11)/6 = 31) log₂ x = - 1/3x = 2^(-1/3)x₁ = 1/∛22) log₂ x = 3x₂ = 2³x₂ = 8 2. log₃ (3^x−8 )= 2 - x, ОДЗ: 3^x - 8 > 0, 3^x > 8, x > log₃ 83^x - 8 = 3^(2 - x)3^x - 8  = 9*(1/3^x)3^(2x) - 8*(3^x) - 9 = 0Пусть 3^x = zz² - 8z - 9 = 0z₁ = -1z₂ = 91)  3^x = - 1, не имеет смысла2)  3^x = 9 3^x = 3²x = 2