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

Question

Considering ${\sqrt{x}}=y$, what is the resolution process of $x^{\sqrt{x}}=\sqrt{x^x}$, with results S={1,4}?

Answer 1

Square both sides to get

$$x^{2\sqrt x}=x^x$$

Either $x=1$, or $2\sqrt x=x\implies x=4$.

Answer 2

Take logs of both sides to get:

$\sqrt x \log x = \frac 12 x\log x$

which has solutions of either $\log x = 0 \implies x = 1$

or:

$\sqrt x = \frac 12 x$

which has solutions of $x = 0,4$. Exclude $x=0$ as it is inadmissible in the original equation, leaving the solutions $x = 1,4$.

Answer 3

Assuming we are only consider real numbers here.

$$x^{\sqrt{x}}=\sqrt{x^x}$$ Firstly, notice that $x\ge 0$, in order for $\sqrt{x}$ to make sense. Also $0^0$ is not defined, thus $x\ne 0$, that is, $x >0$.

Now square both sides:

$$x^{2\sqrt{x}}=x^x$$ Take $\log$ on both sides: $$2\sqrt{x}\log x=x \log x$$

Thus we have:

$$4 x = x^2$$ Or $$\log x = 0$$

So $x = 1$ or $x=4$

Answer 4

Its natural to assume $x> 0$. So let's square both sides to get

$$x^{2\sqrt x}=x^x.$$

Its clear, that $x^x>0$, so use the logarithm to obtain

$$ 2\sqrt x \cdot\log(x)=x \cdot \log(x).$$

Here, you have to exclude $x=1$. Simple substitution of $x=1$ reveals that this would already be a solution. Divide by $\log(x)$. It remains

$$2\sqrt x=x$$

which only has the solution $x=4$ for positive $x$.