Log5 (4x+1)=2-log5(2x+3)

ОДЗ:
$\left \{ {{4x+1\ \textgreater \ 0} \atop {2x+3\ \textgreater \ 0}} \right. ~~~ \left \{ {{4x\ \textgreater \ -1} \atop {2x\ \textgreater \ -3}} \right. ~~~ \left \{ {{x\ \textgreater \ -0,25} \atop {x\ \textgreater \ -1,5}} \right. ~~~\Rightarrow~~~x\in(-0,25;+\infty)$

Решение:
$log_5 (4x+1)=2-log_5(2x+3) \\ log_5 (4x+1)+log_5(2x+3) =2 \\ log_5((4x+1)(2x+3))=log_525 \\ (4x+1)(2x+3)=25 \\ 8x^2+12x+2x+3-25=0 \\ 8x^2+14x-22=0|:2 \\ 4x^2+7x-11=0 \\ D=49-4*4*(-11)=49+176=225 \\ x_1= \frac{-7+15}{8} = \frac{8}{8} =1 \\ \\ x_2=\frac{-7-15}{8} = -\frac{22}{8} =- \frac{11}{4} =-2,75$

$x_2$ - не входит в ОДЗ.

Ответ: 1