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

Решение1)  log₃ (5x + 3) = log₃ (7x + 5)ОДЗ: 5x + 3 > 0, x > - 3/57x + 5 > 0, x > - 5/7x∈ (- 3/5; + ∞)5x + 3 = 7x + 55x - 7x = 5 - 3- 2x = 2x = - 1 -1 ∈ (- 3/5; + ∞)Ответ: - 1 2) log₇ (x - 1) log₇ (x) = log₇ (x)ОДЗ: x - 1 > 0, x > 1x > 0x ∈ (1 ; + ∞)log₇ (x - 1) log₇ (x) - log₇ (x) = 0log₇ (x) (log₇ (x - 1) - 1) = 01)  log₇ (x) = 0x = 7⁰x = 1 1 ∉ (1 ; + ∞)2)  log₇ (x - 1) - 1 = 0log₇ (x - 1) = 1x - 1 = 7¹x = 7 + 1x = 8 8 ∈ (1 ; + ∞)Ответ: x = 8