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Following is the equation

$$x^{x\sqrt x}=x\sqrt{x}$$

We need to find $x$. Please help.

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closed as off-topic by Morgan Rodgers, José Carlos Santos, Paul Frost, Shailesh, YuiTo Cheng May 18 at 2:25

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  • $\begingroup$ We have $$x^{x^{1+1/2}}=x^{3/2}$$ $\endgroup$ – lab bhattacharjee May 17 at 18:28
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    $\begingroup$ $$x=1$$ is one solution. $\endgroup$ – Dr. Sonnhard Graubner May 17 at 18:30
  • $\begingroup$ $x=0$ is, amazingly, not a solution. $\endgroup$ – The Count May 18 at 0:17
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. $\endgroup$ – dantopa May 18 at 1:35
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HINT

Take logs to get $$ x\sqrt{x} \ln x = \ln x + \frac12 \ln x \\ \ln x \left(x \sqrt{x} - \frac32\right) = 0 $$

Can you finish?

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  • $\begingroup$ @user541396 usually one thinks of $0^0$ as undefined $\endgroup$ – gt6989b May 19 at 5:19
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What if $x=0$?

For finite non-zero $x,$

$$x^{x^{3/2}-3/2}=1$$

From Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$, check if $u^m=1$

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