[tex]g(x) = \ln ^2 (x)[/tex]
Recall that if you compose a function to its inverse function, the result will be an identity function. An identity function is just a function where if you input a value, it's just going to ouput that same value. So basically it's just [tex]f(x) = x[/tex].
We are asked to solve for [tex]g(x)[/tex] where [tex]f \circ g(x) = g \circ f(x)[/tex], and [tex]g \circ f(x) = x[/tex]. Notice that [tex]g \circ f(x) = x[/tex] is just an identity function and a composite of [tex]f[/tex] and [tex]g[/tex]. So, it must be [tex]f[/tex] and [tex]g[/tex] must be inverses of each other ([tex]g(x) = f^{-1}(x)[/tex]).
Please refer to my Answer from this Question to know more about Inverse Functions: brainly.com/question/24619467
Please refer to my Answer from this Question to know the needed logarithmic properties to solve: brainly.com/question/24688599
Recall:
[tex]\ln(a) = \log_e (a)[/tex]
Let [tex]y = f(x)[/tex] so that the inverse of [tex]y[/tex] will be [tex]f^{-1}(x)[/tex]
[tex]y = e^{\sqrt x} \\ x = e^{\sqrt y} \\ \ln (x) = \sqrt y \\ \sqrt y = \ln (x) \\ (\sqrt y)^2 = (\ln (x))^2 \\ y = \ln ^2 (x)[/tex]
[tex]f^{-1}(x) = \ln ^2 (x)[/tex]. Since [tex]g(x) = f^{-1}(x)[/tex], we can write it as [tex]g(x) = \ln ^2 (x)[/tex] as well.
