Question

log^5x-log^5 35 < log^5 1/7

Answer 1

5/35=36/х
5х=1260
х= 252

Answer 2

log₅x-log₅35<log₅(1/17)
log₅x<log₅35+log₅(1/17)
log₅x<log₅(35*(1/17))
log₅x<log₅5
основание логарифма а=5, 5>1. знак неравенства не меняем
$left { {{x textgreater 0} atop {x textless 5}} right.$

x∈(0;5)