Find $y$ for $y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$ where $y$ is a positive real number

Question

Here is the answer in the textbook with which I disagree / don't understand: $$y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$$ $$y^{y^{\frac{3}{2}}} = (y^{\frac{3}{2}})^y = y^{\frac{3}{2}y}$$ Comparing indices, we get: $$y^{\frac{3}{2}} = \frac{3}{2}y$$ from $y^{\frac{1}{2}} = {\frac{3}{2}}$, we get: $$y = {\frac{9}{4}}$$ Here what I came up with: $$y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$$ $$(\sqrt[y]{y})^y = (y^{\frac{1}{y}})^y = y$$ $$y^{\sqrt[y]{y}} = y \Rightarrow \sqrt[y]{y} = 1$$ And I get stuck somewhere around: $y = 1^y$

Answer 1

For positive $y$ you have $(\sqrt[y]{y})^y = y$

so $y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y\implies y^{\sqrt[y]{y}} = y \implies y^{\sqrt[y]{y}-1} = 1$

so either $y=1$ or $\sqrt[y]{y}-1=0$

in both cases implying $y=1$

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Alternatively if the original question is actually $y^{y \sqrt{y}} = (y \sqrt{y})^y$

Then this indeed gives $y^{y^{3/2}}=y^{3y/2}$

so either $y=1$ or $y^{3/2}=\frac{3y}{2}$

i.e. either $y=1$ or $y^{1/2}=\frac{3}{2}$

i.e. either $y=1$ or $y=\frac{9}{4}$

Answer 2

$$y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y\implies \sqrt[y]{y}\ln y=y\ln \sqrt[y]{y}=\ln y\implies (\sqrt[y]{y}-1)\ln y=0\implies y=1 $$