Помогите решить уравнение
x=12
Log3(log4(log²3(x-3)))=0, О.Д.З. х-3>0, x>3, log3(log4(log²3(x-3)))=log3(1), log4(log3²(x-3))=1, log4(log²3(x-3))=log4(4), log²3(x-3)=4, log3(x-3)=±2 1)log3(x-3)=2 2)log3(x-3)=-2 log3(x-3)=log3(9) log3(x-3)=log3(1/9) x-3=9 x-3=1/9 x=12 x=28/9