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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?

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

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  • $\begingroup$ @Travis is this still off topic? $\endgroup$ – Ravi Prakash Jul 8 '17 at 18:36
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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 $\{1,4\}$.

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

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  • $\begingroup$ Yes I missed that $\endgroup$ – Atul Mishra Jun 30 '17 at 9:43

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