1)log3(5x+3)=log3(7x+5) 2)log7(x-1)log7x=log7x

Решение
1)  log₃ (5x + 3) = log₃ (7x + 5)
ОДЗ: 5x + 3 > 0, x > - 3/5
7x + 5 > 0, x > - 5/7
x∈ (- 3/5; + ∞)
5x + 3 = 7x + 5
5x - 7x = 5 - 3
- 2x = 2
x = - 1 
-1 ∈ (- 3/5; + ∞)
ответ: - 1 

2) log₇ (x - 1) log₇ (x) = log₇ (x)
ОДЗ: x - 1 > 0, x > 1
x > 0
x ∈ (1 ; + ∞)
log₇ (x - 1) log₇ (x) - log₇ (x) = 0
log₇ (x) (log₇ (x - 1) - 1) = 0
1)  log₇ (x) = 0
x = 7⁰
x = 1 
1 ∉ (1 ; + ∞)
2)  log₇ (x - 1) - 1 = 0
log₇ (x - 1) = 1
x - 1 = 7¹
x = 7 + 1
x = 8 
8 ∈ (1 ; + ∞)
ответ: x = 8