Помогите плиз [latex]log _{ \frac{1}{3} } (x- \frac{1}{6} )=1-log _{ \frac{1}{3} } (x+ \frac{1}{2} )[/latex]

[latex] log_{ \frac{1}{3} }(x- \frac{1}{6} ) =1- log_{ \frac{1}{3} } (x+ \frac{1}{2} )[/latex] ОДЗ: [latex] \left \{ {{x- \frac{1}{6}\ \textgreater \ 0 } \atop {x+ \frac{1}{2}\ \textgreater \ 0 }} \right. , \left \{ {{x\ \textgreater \ \frac{1}{6} } \atop {x\ \textgreater \ - \frac{1}{2} }} \right. =\ \textgreater \ x\ \textgreater \ \frac{1}{6} [/latex] [latex] log_{ \frac{1}{3} } (x- \frac{1}{6} )+ log_{ \frac{1}{3} }(x+ \frac{1}{2} ) =1[/latex] [latex] log_{ \frac{1}{3} } ((x- \frac{1}{6} )*(x+ \frac{1}{2} ))=1[/latex] [latex](x- \frac{1}{6} )*(x+ \frac{1}{2} )=( \frac{1}{3} ) ^{1} [/latex] [latex] x^{2} - \frac{1}{6} x+ \frac{1}{2} x- \frac{1}{12}= \frac{1}{3} |*12[/latex] 12x²+4x-5=0 x₁=-5/6 посторонний корень x₂=1/2 ответ: х=0,5