log0.3(2x-4)>log0.3(x+1)

Question

log0.3(2x-4)>log0.3(x+1)

Answer 1

log{0.3}(x+1)\\2x-4>0 ||x+1>0\\x>2||x>-1\\\\a=0,3 (<1)\\2x-4<x+1\\x<5\\\\\boxed{x\in(2;5)}< var="">' alt='log_{0.3}(2x-4)>log{0.3}(x+1)\\2x-4>0 ||x+1>0\\x>2||x>-1\\\\a=0,3 (<1)\\2x-4<x+1\\x<5\\\\\boxed{x\in(2;5)}< var="">' align="absmiddle" class="latex-formula">

Answer 2

ОДЗ: 2x-4>0

x>2

x+1>0

x>-1

 

основания равны приравниваем.

2x-4

2x-x<1+4</p>

x<5</p>

учитывая ОДЗ:

2

 

Ответ: 2