Question 1017935
If a function is one-to-one, it has a unique inverse function. The inverse of an expential function is a logarithmic function, so it cannot be itself.


*[tex \large f^{-1}(x) = \frac{1}{2} \ln x] is correct. To check, you can simply compute *[tex \large f^{-1}(f(x))]


*[tex \large f^{-1}(f(x)) = f^{-1}(e^{2x}) = \frac{1}{2} \ln (e^{2x}) = \frac{1}{2} \cdot 2x = x]


Similarly, *[tex \large f(f^{-1})(x) = x]. So they are inverses.