Log3 (5-4x)= log3 (x-1) log3 (7-x)> 1 log1/3 (3-5x)< -3 решить неравенство?

Log₃ (5 - 4x) = log₃ (x - 1)
ОДЗ: 5 - 4x > 0, x < 5/4
x - 1 > 0, x > 1 
ОДЗ: 1  < x < 5/4, 1 < x < 1,25 
5 - 4x = x - 1
- 4x - x = - 1 - 5
-5x = - 6
x = 6/5
x = 1,2
2) log₃ (7 - x) > 1
ОДЗ: 7 - x > 0, x < 7
3 > 1
7 - x > 3
- x > 3 - 7
-x > - 4
x < 4
 x ∈ (- ≈ ; 4)
3) log₁/₃ (3 - 5x) < - 3
0 < 1/3 < 1
3 - 5x > (1/3)⁻³
-5x > 3³ - 3
- 5x > 27 - 3
- 5x > 24
x < - 4,8