Need help finding inverse under $a\circ b = a^b b^a$ I'm going through some problems in Theorems, Corollaries, Lemmas, and Methods of Proof, and I'm stuck at a certain problem that seemed very interesting until I couldn't solve it for the life of me.
Let $\circ$ be defined on $\mathbb R^+$ as $a\circ b = a^b b^a$. One can show that $\circ$ is Abelian, that $\mathbb R^+$ is closed under $\circ$, and that there exists identity element $e=1$ such that $a\circ e = e \circ a = a$.
The next part of the problem asks to find $2^{-1}$, which means solving for $x$ where $2^x x^2 = 1$. Because $\circ$ is Abelian, there's no other way to arrange this expression. I tried taking logarithm of both sides to get
$$\begin{align}
\log_2(2^x x^2) &= \log_2(1) \\
\log_2(2^x) + \log_2(x^2) &= 0 \\
x + 2\log_2(x) &= 0 \\
x &= -2\log_2(x) \\
1 &= -2 \frac{\log_2(x)}{x} \\
-\frac{1}{2} &= \log_2(x^{\frac{1}{x}}) \\
2^{-\frac{1}{2}} &= x^{\frac{1}{x}}
\end{align}$$
This doesn't really get me anywhere. I tried other manipulations, but all expressions I got were quite ugly. WolframAlpha doesn't return any meaningful results for positive reals. I keep coming back to this problem, but it looks like I won't be able to solve it. Perhaps I'm missing some algebra I have to use to get the result.
 A: The equation $2^{-x}=x^2$ admits three real solutions : $x=-2,\;x=-4\;$ 
and a more sophisticated solution involving the LambertW function as returned by WolframAlpha (the only positive solution will be this last one).
These solutions may be obtained by writing $\,2^{-x}=x^2\,$ as $\,1=x\,e^{x\frac{\ln(2)}2}\,$ or $\,-1=x\,e^{x\frac{\ln(2)}2}\,$ that we may express as :
$\;\displaystyle \frac{\ln(2)}2=\left(x\frac{\ln(2)}2\right)\,e^{\left(x\frac{\ln(2)}2\right)}\;$ or $\;\displaystyle \frac{-\ln(2)}2=\left(x\frac{\ln(2)}2\right)\,e^{\left(x\frac{\ln(2)}2\right)}$
Since the LambertW function is defined implicitly by $\displaystyle z=W(z)e^{W(z)}$ we see that the answers are given by $\;W\left(\frac{\ln(2)}2\right)=x\frac{\ln(2)}2\;$ and $\;W\left(\frac{-\ln(2)}2\right)=x\frac{\ln(2)}2\;$ that is by
$$\tag{1}\boxed{\displaystyle x=\frac2{\ln 2} W\left(\frac{\ln 2}2\right)}$$ and $$\tag{2}\;\boxed{\displaystyle x=\frac2{\ln 2} W\left(\frac{-\ln 2}2\right)}$$
The subtle point is that this second solution $(2)$ will be split in two sub-solutions since the $\rm LambertW$ function admits two branches for $x \in \left(-\dfrac1e,0\right)\;$ (i.e. for $ W\left(\dfrac{-\ln 2}2\right)$) : 


*

*the upper one gives $W_0\left(\dfrac{-\ln 2}2\right)=-\ln 2\;$ (divided by $\ln 2$ and multiplied by $2$ returns the $x=-2\,$ solution)  while 

*the lower branch gives the value noted $W_{-1}\left(\dfrac{-\ln 2}2\right)$ (after multiplication by $\dfrac2{\ln 2}$ this returns the $x=-4\;$ solution).


The solution $(1)$ is the only real and positive solution :
$$\boxed{\displaystyle x=\frac2{\ln 2} W\left(\frac{\ln 2}2\right)\approx 0.766664695962123}$$
To clarify further : you probably had to find an approximation of the answer (possibly using the plot of $x\to 2^{-x}$ and $x \to x^2$ on the same graphics as shown on WolframAlpha) or find it numerically (say by iterations or with a Taylor series) or play with the LambertW function... 
