Solve the equation `log_(2)(3-x)-log_(2)(("sin"(3pi)/4)/(5-x)}=1/2+log_(2)(x+7)`

Explanation

This equation is equivalent to
`log_(2)(3-x)=log_(2)(("sin"(3pi)/4)/(5-x))+1/2log_(2)2+log_(2)(x+7)`
`implieslog_(2)(3-x)=log_(2)(1/(sqrt(2)(5-x)))+log_(2)+log_(2)(x+7)`
which is equivalent to the system
`{(3-xgt0),(1/(sqrt(2)(5-x))gt0),(x+7gt0),((3-x)=(sqrt(2)(x+7))/(sqrt(2)(5-x)):}`
`implies{(xlt3),(xlt5),(xgt-7),((x-1)(x-8)=0:}`
Hence `x_(1)=1` is only root of the original equation.