Решите уравнение
Log3 (3x^2) * log3 x = 1 одз x>0 log ab = log !a ! + log !b! (log3 3 + log3 x^2)*log3 x =1 (1 + 2log3 x)*log3 x = 1 2log3^2 x + log3x - 1=0 log3 x =t 2t^2+t-1=0 D=1+8=9 t12=(-1+-3)/4= - 1 1/2 log3 x=-1 x=1/3 log3 x=1/2 x=√3