[tex] log_{3}9+ log_{3}( 2^{x}-2)= log_{3}2+ log_{3}( 4^{x-1}-1) \\ 9*( 2^{x}-2)=2( 4^{x-1}-1) \\ fie:2^x=t,t\ \textgreater \ 0 \\ 9(t-2)=2(t^2* \frac{1}{4}-1) \\ t^{2}-18t+32=0 \\ t=2;t=16 \\ revenim \\ 2^x=2;x=1 \\ 2^x=16;x=4[/tex]
x=1,nu poate solutie ,deoarece DVA:x>1
S={4}
