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Taking $\ln$ both sides, we have that ${%Error. "additional" is a bad command. \displaystyle x\ln x=\frac{1}{2}\cdot\ln \frac{1}{2}} \\\Rightarrow \frac{x}{\frac{1}{2}}=\frac{\ln\frac{1}{2}}{\ln x}$ but since $a\geq b \Rightarrow \ln a\geq \ln b$, we must have one of $\displaystyle \max \left\{\frac{x}{\frac{1}{2}},\frac{\ln\frac{1}{2}}{\ln x}\right\}\geq 1 \geq \min \left\{\frac{x}{\frac{1}{2}},\frac{\ln\frac{1}{2}}{\ln x}\right\}$, so we must have $x=\frac{1}{2}$

The solution of $campos$ is not sufficient ! It's easy to see that $x_1=\frac 14,\ x_2=\frac 12$ are solutions . There are not other solutions because of the function $f(x)=x^x$ is decreasing on $]0,\frac 1e]$ and increasing on $[\frac 1e,+\infty[$

i thought this can't be so easy..., isn't $f(x)=\ln x$ strictly increasing on $\mathbb{R^+}$?

i think i've found a solution... hope it's correct this time... as in the previous post we have $x^x=\frac{1}{\sqrt 2} \Leftrightarrow x \ln x= \frac{1}{2}\ln \frac{1}{2}$ now, let $x=\left(\frac{1}{2}\right)^k$, then, $x \ln x= \frac{1}{2}\ln \frac{1}{2} \Rightarrow \left(\frac{1}{2}\right)^k\ln \left(\frac{1}{2}\right)^k=\frac{1}{2}\ln \frac{1}{2} \\\\\Rightarrow 2^k=2k$ but, since $f(x)=2^x$ is an exponential function and $g(x)=2x$ is a linear function, we must have $f(x)$ and $g(x)$ intersect in at most two points, namely $k=1$ and $k=2$... so we must have $x=\frac{1}{2}$ and $x=\frac{1}{4}$ are the solutions to the equation.

How do you know that $x$ must be a power of $1/2$?

As I see it, $k$ is not necessary an integer...

Derivating, we have that $(x^x)’ = x^x(\ln x + 1)$. Therefore, $x^x$ is decreasing when $0 < x < \frac{1}{e}$ and increasing when $x > \frac{1}{e}$. It follows that there can be at most 2 solutions. After some trial and error, we see that $\frac{1}{4}$ and $\frac{1}{2}$ are solutions and by the aforementioned reasoning, they are the only solutions. $\blacksquare$

We can see that $x^x$ is decreasing when $0 < x < \frac{1}{e}$, and increasing when $x> \frac{1}{e}$, I used trial and error, to get that $\frac{1}{4}$ and $\frac{1}{2}$, are the only solutions. $\square$