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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

(x^a )^b =x^(ab)   (x^a )^a =x^a^2    ⇒ g(x)=x^(√2)

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

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

very nice!  are there any other possibilities?

$${very}\:{nice}! \\ $$$${are}\:{there}\:{any}\:{other}\:{possibilities}? \\ $$

Commented by MJS_new last updated on 30/Apr/21

I′m not sure  but (x^(√2) )^(√2) =x^2  is only true for  x=re^(iθ) ∧r∈R^+ ∧−((π(√2))/2)≤θ<((π(√2))/2)

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

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