(log3(x))^2 -7log3(x)+12=0
(㏒3(х))²-7㏒3(х)+12=0 ㏒3(х)=t t²-7t+12=0 t1+t2=7 t1*t2=12 t1=3 ㏒3(x)=3 x=3³ x1=27 t2=4 ㏒3(x)=4 x=3^4 x2=81 Ответ: х1=27 х2=81