Log(x+1)+log(2x+1)=1

$log_6 (x+1)+log_6 (2x+1)=1$

0;2x+1>0" alt="x+1>0;2x+1>0" align="absmiddle" class="latex-formula">

-1;2x>-1" alt="x>-1;2x>-1" align="absmiddle" class="latex-formula">

-1;x>-0.5" alt="x>-1;x>-0.5" align="absmiddle" class="latex-formula">

=-0.5" alt="x>=-0.5" align="absmiddle" class="latex-formula">

$log_6 ((x+1)(2x+1))=log_6 6$
$(x+1)(2x+1)=6$
$x^2+x+2x+2=6$
$x^2+3x+2-6=0$
$x^2+3x-4=0$
$(x+4)(x-1)=0$
$x+4=0;x_1=-4<-0.5$ - не подходит
$x-1=0;x_2=1$
ответ: 1