Одно обстоятельство требует уточнения. Log (14-4x)=log4(2x+2)
log 1/6(7x-9)=log1/6x

To solve these equations, we will first rewrite them in exponential form by setting the quantities inside the logarithms equal to each other.

For the first equation:
14-4x = 4(2x+2)
14-4x = 8x + 8
14 - 8 = 8x + 4x
6 = 12x
x = 1/2

For the second equation:
1/6(7x-9) = 1/6x
Multiplying both sides by 6 to get rid of the fractions:
7x - 9 = x
6x = 9
x = 3/2

So, the solutions to the equations are:
x = 1/2 for the first equation
x = 3/2 for the second equation.
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log4(14-4x) = log4(2x+2)

14 - 4x = 2x + 2

14 - 2 = 2x + 4x

12 = 6x

x = 2

Ответ: 2

log (1/6)(7x-9) = log(1/6)x

7x - 9 = x

7x - х = 9

6x = 9

x = 9/6 = 3/2 = 1,5

Ответ: 1,5