Question

log in baza 2 din [(2^(-x+1) +1] = x

Answer 1

$log_{2} (2^{-x+1}+1) =x \\ 2^{-x+1}+1= 2^{x} \\ 2^{-x}-2^x+1=0 \\ fie 2^x=t;t\ \textgreater \ 0 \\ \frac{1}{t}-t+1=0 \\ t^2-t-1=0 \\ delta=5 \\ t_{1}= \frac{1- \sqrt{5} }{2}; t_{2}= \frac{1+ \sqrt{5} }{2} \\ revenim:2^x= \frac{1+ \sqrt{5} }{2};x= log_{2} \frac{1+ \sqrt{5} }{2}$