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Question Number 139672 by mathlove last updated on 30/Apr/21

Answered by MJS_new last updated on 30/Apr/21

$${\left(x^a\right)}^b=x^{ab}$$$${\left(x^a\right)}^a=x^{a^2}$$$$\Rightarrow g\left(x\right)=x^{\sqrt{2}}$$

Commented by mr W last updated on 30/Apr/21

$$verynice!$$$$arethereanyotherpossibilities?$$

Commented by MJS_new last updated on 30/Apr/21

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}$$$$\mathrm{but}{\left(x^{\sqrt{2}}\right)}^{\sqrt{2}}=x^2\mathrm{is}\mathrm{only}\mathrm{true}\mathrm{for}$$$$x=r{\mathrm{e}}^{\mathrm{i}\theta }\wedge r\in {\mathbb{R}}^{+}\wedge -\frac{\pi \sqrt{2}}{2}⩽\theta <\frac{\pi \sqrt{2}}{2}$$

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