Question

Log2(2^(-x+1)+1)=x cum se rezolva?

Answer 1

[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]