Question

$log_{2}( 4^{x}+4)=x+log_{2}(-3+2^{x+1})$

Answer 1

$log_{2} (4^x+4)= log_{2}2^x+ log_{2}(-3+ 2^{x+1}) \\ 4^x+4=2^x(-3+ 2^{x+1}) \\ fie:2^x=t,t\ \textgreater \ 0 \\ t^{2}+4 =t(-3+2t) \\ t^{2}+4=-3t+2 t^{2} \\ t^{2}-3t-4=0 \\ t=-1;t=4 \\ revenim.la.substitutie \\ 2^x=4 \\ x=2$
S={2}