Помогите,пожалуйста))

$log_{6}\frac{16-4x}{7}=log_{6}\frac{1}{35}$

$\left \{ {{\frac{16-4x}{7}=\frac{1}{35}} \atop {\frac{16-4x}{7}\ \textgreater \ 0}} \right.$

$\left \{ {35*(16-4x)=7} \atop {x\ \textless \ 4}} \right.$

$\left \{ {560-140x=7} \atop {x\ \textless \ 4}} \right.$

$\left \{ {140x=560-7} \atop {x\ \textless \ 4}} \right.$

$\left \{ {x=3.95} \atop {x\ \textless \ 4}} \right.$

Ответ: х=3.95