I think you are making hard work of it
the inverse of
\(y=5^{logx}\)
is simply
\(x=5^{logy}\)
You just have to switch the x and y over.
there would be restrictions on x and on y but the question isn't worrying about that.
Here is the graphs
https://www.desmos.com/calculator/4ftfa2bny7
See they are reflections of each other about y=x
The inverse function can be graphed by: \(y=x^{\frac{1}{\log5}}\)
Here's a graph:
https://www.desmos.com/calculator/aijpzpouk1
In other words...
If \(f(x)=5^{\log x}\) then the inverse is \(f^{-1}(x)=x^{\frac{1}{\log 5}}\)
And...
\(f(f^{-1}(x))\ =\ f(x^{\frac{1}{\log 5}})\ =\ 5^{\log(x^{\frac{1}{\log 5}})}\ =\ 5^{\frac{\log x}{\log 5}}\ =\ 5^{\log_5 x}\ =\ x\)
The options leave it in the form that is solved for x, and so I left it like that to match the options