помогите решить пжл а) (1/2)^x=8б)log3 х=2в)3^2x+1=27г) log1/3 (4x+1)= - 2

1) (1/2)^x=8 --> x=log(1/2,8) = -3

2) log(3,x)=2 --> x = 3^2 = 9

3) log(3, 27) = 2x+1 --> x=(3-1)/2 = 1

4) log(1/3, 4x+1) = -2 --> 4x+1 = 9 --> x = 2