Here is another method:
[tex]0.5 log_3 (x) = 2[/tex]
Multiply both sides by 10 to get rid of the decimal:
[tex]10 * 0.5 log_3 (x) = 10 * 2[/tex]
[tex]5 log_3 (x) = 20[/tex]
Divide 5 out
[tex]\frac{5 log_3 x}{5} = \frac{20}{5}[/tex]
[tex]log_3 x = 4[/tex]
Now use the log rule [tex]a = log_b b^a[/tex]
[tex]4 = log_3 (3^4) = log_3 (81)[/tex]
Since we have the same base
[tex]log_3 (x) = log_3 (81)[/tex]
x = 81