Решите уравнение log₈(3-2x)-log₈(4-3x)=log₈ 2

Log₈(3-2x)-log₈(4-3x)=log₈2
ОДЗ:
$\left \{ {{3-2x\ \textgreater \ 0} \atop {4-3x\ \textgreater \ 0}} \right. , \left \{ {{-2x\ \textgreater \ -3} \atop {-3x\ \textgreater \ -4}} \right. , \left \{ {{x\ \textless \ \frac{3}{2} } \atop {x\ \textless \ \frac{4}{3} }} \right. , \left \{ {{x\ \textless \ 1,5} \atop {x\ \textless \ 1 \frac{1}{3} }} \right.$
=> $x\ \textless \ 1 \frac{1}{3}$
$log_{8} \frac{3-2x}{4-3x} = log_{8} 2$
$\frac{3-2x}{4-3x} =2$
3-2x=2*(4-3x)
-2x+6x=8-3
4x=5
x=1,25
$1,25 \ \textless \ 1 \frac{1}{3}$
=> x=1,25 - корень уравнения