Question

3+$log_{ \frac{1}{2} }( \frac{32}{x+4})=log_{ \frac{1}{2} } (1-x)$

Answer 1

$log_{ \frac{1}{2} } \frac{1}{8}+ log_{ \frac{1}{2} } \frac{32}{x+4} = log_{ \frac{1}{2} }(1-x) \\ \frac{1}{8}* \frac{32}{x+4}=1-x \\ \frac{4}{x+4}=1-x \\ 4=(1-x)(x+4) \\ x^{2} +3x=0 \\ x(x+3)=0 \\ x=0;x=-3$
S={-3;0}