help with yr 8 yearly maths question: ?

81^(1 - 2x) = 1/9(√3)
(3^4)^(1 - 2x) = 1/(3^2)[3^(1/2)]
3^[4(1 - 2x)] = 3^-2[3^(1/2)]
3^[4(1 - 2x)] = 3^(-2 + 1/2)
4(1 - 2x) = -1 1/2
4 - 8x = -1 1/2
-8x = -1 1/2 - 4
-8x = -3/2 - 8/2
-8x = -11/2
x = (-11/2)/-8
x = (-11/2)(-1/8)
x = 11/16

Question is unclear.

Is it 1 / ( 9√3 )????? OR ( 1 / 9 ) ( √3 ) ???????

Two different things.

The homework is to allow you learn how to rearrange functions and arithmetics of exponentials.

1) 81=9^2

Therefore: 81^(1-2x)= 9^(2(1-2x))=9^(2-4x)

2) 1/9root3= 9^(-1/3)

Now Put back into original equation

9^(2-4x)=9^(-1/3)

2-4X= -1/3

x=7/12

81^(1-2x) = 3^1/9

3^4-8x = 3^1/9

4-8x = 1/9

-8x=1/9-4

x = 35/72

x=0.486

81^(1-2x)=1/9root3
log 81^(1-2x)=log(1/9root3)
(1-2x) log(81) = log(1/9 root3)
1-2x = log(1/9 root3)/log(81)
x = (1- log(1/9 root3)/log(81) )/2

x = 0.6875