# What is the value of $x$ in this equation using logarithms [closed]

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?

## closed as off-topic by Travis, Aretino, kingW3, Namaste, Antonios-Alexandros RobotisJun 30 '17 at 12:19

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• @Travis is this still off topic? – Ravi Prakash Jul 8 '17 at 18:36

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 $\{1,4\}$.
• Which implies two solutions: $x=4$ or $x=1$. – freakish Jun 30 '17 at 9:35
• Arguably $0$ may also be a solution as the right limit of both expressions is $1$ as $x \to 0^+$ – Henry Jun 30 '17 at 9:46
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$$x=1,4$$