How can I find the inverse function of $f(x) = x^x$? I cannot seem to find the inverse of this function, or any function in which there is both an $x$ in the exponent as well as the base. I have tried using logs, differentiating, etc, etc, but to no avail.
[Math] Inverse function of $x^x$
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If the inverse exists, you just write $y=x^2-6x+11$ and use the quadratic formula to get $x$ in terms of $y$. To see if it exists, you need to ensure that for a given $y$ there is only one $x$. The obvious threat is the $\pm$ sign in the quadratic formula.
First of all, you need the function to be bijective (that is, injective and surjective) to be able to find an inverse. For example, $f:\Bbb R\ni x \mapsto 1$ has no inverse.
Then, you need to understand what functions are. A function $f$ is $(D,C,G)$ where $G\subset D \times C$. $D$ is the domain, $C$ is the codomain and $G$ is the graph of $f$, a subset of $D\times C$ with the property that $\forall x \in D, \exists ! y \in C, (x,y)\in G$. Because of this property, we can assign an $y\in C$ to any given $x\in D$ and so we decide to give a name to that $y$: $f(x)$.
Saying that $f$ is injective is saying that $\forall x_1,x_2\in D, f(x_1)=f(x_2)\implies x_1=x_2$. And saying that it is surjective means that $\forall y \in C, \exists x \in D, f(x)=y$. Now if you translate those properties in terms of graph, you get $\forall x_1,x_2\in D, \forall y \in C, (x_1,y)\in G \land (x_2,y)\in G \implies x_1=x_2$ and $\forall y \in C,\exists x \in D, (x,y)\in G$. The first one tells you that for $y\in C$, there is at most one $x\in D$ so that $(x,y)\in G$ and the second one tells you that there is at least one so there is exactly one. Which means $\forall y\in C,\exists !x \in D, (x,y)\in G$. This looks very similar to the property about graph of functions. In fact, this means that $(C, D, \{(y,x)\mid (x,y)\in G\})$ is a function. We'll call it $f^{-1}$.
This means that given a bijective function $f$, we already know $f^{-1}$. But since we tend to dislike writing infinite sets, we prefer giving a function by its domain $D$, its codomain $C$ (which we denote by $f:D\to C$) and then a formula that tells you how to find $f(x)$ in terms of $x$. But not all functions can be expressed that way. For example, you write $f:\begin{array}{l}\Bbb R_+\to \Bbb R_+\\x\mapsto \sqrt{x}\end{array}$ . But how do you define $\sqrt{x}$? Well in fact, you started with $g:\begin{array}{l}\Bbb R_+\to \Bbb R_+\\x\mapsto x^2\end{array}$, showed it was bijective so $g^{-1}$ existed. And then, since it looked like a really important function, we decided to give it its own symbol and write $\sqrt{x}$ instead of $g^{-1}(x)$.
In some cases, if you express $f$ in terms of already known functions, $f^{-1}$ will also be expressible in terms of already known function but it is very rare. And so when we find an interesting inverse function, we give it its own name. Examples are $\ln =\exp^{-1}$, $\sqrt{\cdot}=(\cdot^2)^{-1}$ and trigonometric inverse functions ( http://en.wikipedia.org/wiki/Inverse_trigonometric_functions ).
Best Answer
As the other user mentioned, it is basically the application of Lambert W Function.
Say, $x^x = z$ which implies, $x \ln x = \ln z$.
Now, I can write: $x = e^{\ln x} $ using the properties of logarithms and exponential functions.
Therefore, $$\ln x = W \ln z \\ x = e^{W \ln z} $$
which is indeed the inverse of $x^x$ .
I suggest you to go through the Wikipedia's page for Applications of Lambert W Function.
Hope it helps!