[tex]log_2(2^{-x+1}+1)=xlog_22=log_22^x=\ \textgreater \
2^{-x+1}+1=2^x\\
2^x= \frac{1}{2^{x-1}}+1\\
2^x= \frac{2}{2^x} +1 \\Notam\ 2^x=t\\
t= \frac{2}{t} +1\\
t^2-t-2=0=\ \textgreater \ \triangle=9\\
t_1=2; t_2=-1\\
2^x=2=\ \textgreater \ x=1
[/tex]
