Question 299184
{{{ log(2,(2-2x))+log(2,(1-x))=5 }}}
{{{ log(2,(2-2x)(1-x))=5 }}}
{{{ (2-2x)(1-x)=2^5 }}}
{{{ 2-2x-2x+2x^2=32 }}}
{{{ 2x^2-4x+2=32 }}}
{{{ x^2-2x+1=16 }}}
{{{ x^2-2x-15=0}}}
{{{ (x+3)(x-5)=0 }}}
.
x = {-2, 5}
Toss out 5 as a solution, it's an extraneous solution leaving (causes problems with the original equation).  Therefore, the solution:
x = -2