[tex]\displaystyle\bf\\log_3(x^2-2x+4)=1\\\\x^2-2x+4=3^1\\\\x^2-2x+4-3=0\\\\x^2-2x+1=0\\\\x^2-2x\cdot1+1^2=0\\\\(x-1)^2=0\\\\x-1=0\\\\x=1~~~(radacina~dubla)\\\\\boxed{\bf x_1=x_2=1}\\\\Verificare:\\\\ log_3(x^2-2x+4)=\\=log_3(1^2-2\times1+4)=\\=log_3(1-2+4)=\\=log_3(3)=1\\Corect[/tex]