TITLE: Does the equation $y = x\log x$ have a solution?
QUESTION [0 upvotes]: Consider the function: $y = x\log x$  
Now, put $y = 1$: 
so, 
\begin{align*} 
x\log x & = 1\\  
\log x^x & = 1
\end{align*}
Now, this clearly gives a contradiction to the mathematical properties, for demonstration, assume that the base of log here is $2$, i.e. $\log = \log_2$.  
So, our equation is:  
$$\log x^x = \log_2 2$$  
For both sides to be equal, the only thing that will make the equations true is when $1 = 2$, which is impossible.
EDIT TO EXPLAIN HOW 1 = 2:
so we reach a point where
 $$\log_2x^x = \log_2 2$$
This suggests that xx must be equal to 2, meaning that the base and power must be equal, and the powers are equal when and only when 1 = 2.
 Where am I making a mistake in understanding?

REPLY [4 votes]: $\log_2 x^x = \log_2 2$
So far so good. By the one-to-one nature of the logarithm function, you can conclude:
$x^x = 2$.
But I fail to see how you conclude that $1=2$?
Edited to attempt to answer the clarified post
$x^x = 2$ does not imply that $1=2$
$x^x = 2$ cannot be solved exactly through elementary means (without invoking special functions such as the Lambert W). However, it can quite easily be solved numerically to give an approximate solution of $x \approx 1.5596$.
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