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

z ² - 8z - 9 = 0

z₁ = - 1

z₂ = 9

1) 3^x = - 1, не имеет смысла

2) 3^x = 9

3^x = 3 ²

x = 2