I am new to logarithms and I need to find out the value(s) of $x$ in the below equation, preferably by logarithms.
$$x^{\sqrt{x}} = (\sqrt{x})^x$$
Edit:
What I had already done before asking this question is: I tried taking logarithm on both sides and got 4 as a solution.
But I need two solutions of this equation. Which is the other solution? How can it be got in a good way?
Suppose $x>0$. Taking the log it is equivalent to $$ \sqrt{x} \log x = x \log\sqrt{x}=\frac{1}{2} x \log x. $$ Note that $x=1$ is a solution. Otherwise $\sqrt{x}\log x \neq 0$. Simplify $\sqrt{x}\log x$ hence $$ 1=\frac{1}{2}\sqrt{x} \implies x=4. $$ Therefore all solutions are $\{1,4\}$.
Note that $1$ is a solution
It is $$x^{x^\frac{1}{2}}=x^{\frac{x}{2}}$$so $$x^{\frac{1}{2}}=\frac{x}{2}$$Taking log $$\frac{1}{2}\log x=\log x-\log 2$$$$\log x=2\log 2$$$$\log x =\log 2^2$$Taking antilog $$x=2^2$$$$x=4$$So$$x=1,4$$
