How to calculate log (base9)= square root of 3

I will thank you so much for this help!

if a^n=b then b^(1/n)=a

(sqrt(3)^2)^2=9
sqrt(3)^4=9
sqrt(3)=9^(1/4) taking log base 9
log(base 9)(sqrt(3))=1/4=.25

There are no variables so we could simplify.
log(base9) [of what] = √3

But I think you mean
log(base 9) of 3 = 1/2
Then 9^(1/2) = 3

x = 9^1/3 ?