Question

log0,7(3x-5) < log0,7(x+1)

Answer 1

$log _{0,7}(3x-5) textless log _{0,7} (x+1)$
Так как $0 textless 0,7 textless 1,$  то
3x - 5 > x + 1                  ОДЗ: 1) 3x - 5 > 0                   2) x + 1 > 0
3x - x > 1 + 5                                3x > 5                            x > - 1
2x > 6                                            x > $1 frac{2}{3}$
x > 3
x ∈ (3 ; + ∞)