<s>[INST] Properties of $\log^2(x)$ I have been trying to figure out the solution to a logarithm problem, and keep running into the equation $\\10\log^2_2(x) = x$. In so doing I've been trying to simplify $\log_2^2(x)$. What I have is
$\log_2^2(x) = \log_2(x)\log_2(x) = \log_2(x^{\log_2(x)})$
which as far as I can tell is no easier to work with.
Am I missing something? Or is there a completely different alternative way to look at this problem? Thanks!
 [/INST] This is a fun one! We can solve it using the Lambert W Function, the inverse of the function
$$f(x) = xe^x$$
$$10\log_2^2(x) = x$$
$$\sqrt{10}\log_2(x) = \sqrt{x}$$
$$2^{\log_2(x)\sqrt{10}} = 2^{\sqrt{x}}$$
$$x^{\sqrt{10}} = 2^{\sqrt{x}}$$
$$x = 2^{\frac{\sqrt{x}}{\sqrt{10}}}$$
$$x = e^{\frac{\sqrt{x}}{\sqrt{10}}\log 2}$$
$$(\sqrt{x})^2 = e^{\frac{\log 2}{\sqrt{10}}\sqrt{x}}$$
$$\sqrt{x} = e^{\frac{\log 2}{2\sqrt{10}}\sqrt{x}}$$
Let $c = \frac{\log 2}{2\sqrt{10}}$:
$$\sqrt{x} = e^{c\sqrt{x}}$$
$$-c\sqrt{x}e^{-c\sqrt{x}} = -c$$
$$-c\sqrt{x} = W(-c)$$
$$\sqrt{x} = \frac{W(-c)}{-c}$$
$$x = \frac{(W(-c))^2}{c^2}$$
$$x = \frac{\left(W\left(-\frac{\log 2}{2\sqrt{10}}\right)\right)^2}{\left(\frac{\log 2}{2\sqrt{10}}\right)^2}$$
$$x =\left(W\left(-\frac{\log 2}{2\sqrt{10}}\right)\right)^2\frac{\log^2 2}{40}$$
There are actually $3$ solutions for $x$. One results in taking this function as is. The other results in using $-\sqrt{10}$ instead of $\sqrt{10}$. The last results in taking $\sqrt{10}$ and using the second branch $W_{-1}$ of the Lambert W function (since there are multiple solutions to $xe^x = y$ for all $y<0$). 
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