relation betwn ln and e If $f(x) = ln(x)$ and $f^-1(x) = e^x$ then is $e^x = 1/ln(x)$???
because I see $e^9 = 8103$ but $1/ln9 = .455$
How are they reverse? I don't understand!

As copper.hat mentioned it, there is a notation "clash" here.
More precisely, let there be $E$ a set and $$\matrix{\star: & E^2 & \to & E\\ & a,b & \mapsto & \star(a, b) & =a \star b}$$
such that $(E, \star)$ is a group.


*

*Either we notate $\star$ like the addition and name $0_E$ the neutral element of E, and $\forall a \in E, ~~-a$ its symmetrical, so that $a\star(-a) = (-a)\star a = 0_E$

*Or we can notate $\star$ like the multiplication and name $1_E$ the neutral element, and $\forall a \in E ~~a^{-1}$ its symmetrical. Here again, $a \star a^{-1} = a^{-1} \star a = 1_E$


It actually makes no difference, but can be useful when manipulating several laws, for instance in rings, fields or *-Algebra.
Now back to the question
The field we use the most is $(\mathbb{R}, +,\times)$
So in most situations we encounter, when seeing something like $-a, a^{-1}$, assuming they are the (respectively) opposite and inverse of a real named $a$ is not that risky.
But the set of all bijections from $E$ to $F$, $A$, comes with a defined law, noted $\circ$ such that : $$\matrix{\circ : & A^2 & \to & A\\ & u,v & \mapsto &\circ(u,v) & = u\circ v : & E & \to & F\\ &&&& & x & \mapsto & (u\circ v)(x) & = u(v(x)) }$$
And, as $(A,\circ)$ turns out to be a group, any bijection of A has a unique symmetrical  in regards of $\circ$.
And as we notate $\circ$ just like multiplication, we notate it $f^{-1}, \forall f \in A$, such that $f\circ f^{-1} = f^{-1}\circ f = 1_A$.
note : if $(E = F), 1_A = id_E : x \mapsto x$
Problem
let $f \in A, x\in E$,
And that's were it get confusing : we know that $f(x)\in F$, so if $(F,\times)$ is a group, when we read $(f(x))^{-1}$, sometimes noted $f^{-1}x$ not to confuse with $fx^{-1}$, we may safely assume that it's the inverse of $f(x)$ in regards of the multiplication defined on $F$.
But now if we have $y \in F, f^{-1}(y)$ now refers to the image of $y$ through $f^{-1} : F \to E$ the inverse function of $f$.
But what now if $E=F=\mathbb{R}$ ? well, being rigorous is the only way.
Solution
First right down the exact sets where your functions are defined. For instance :
$$ \matrix{exp : & \mathbb{R} & \to & \mathbb{R}^{+*}\\ln : & \mathbb{R}^{+*} & \to & \mathbb{R}}$$
You first notice that the output of one is included or equal to the input of the other...
