Question Number 127305 by MathSh last updated on 28/Dec/20
$$\left({a}−\mathrm{3}\right)^{−\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$${the}\:{set}\:{of}\:{possible}\:{values}\:{of}\:{a} \\ $$$$\left.{a}\right)\left(−\infty;+\infty\right) \\ $$$$\left.{b}\right)\left(\mathrm{3};+\infty\right) \\ $$$$\left.{e}\right){a}\neq\mathrm{3} \\ $$
Answered by MJS_new last updated on 28/Dec/20
$$\mathrm{depending}\:\mathrm{on}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{we}\:\mathrm{use}… \\ $$$${b}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{5}}]{{a}−\mathrm{3}}} \\ $$$$\left(\mathrm{1}\right)\:{a}\in\mathbb{R}\wedge{b}\in\mathbb{R};\:\sqrt[{\mathrm{5}}]{{x}}=\mathrm{sign}\:{x}\:\sqrt[{\mathrm{5}}]{\mid{x}\mid}\:\left(\mathrm{i}.\mathrm{e}.\:\sqrt[{\mathrm{5}}]{−\mathrm{32}}=−\mathrm{2}\right. \\ $$$$\Rightarrow\:{a}\neq\mathrm{3}\:\Rightarrow\:{a}\in\mathbb{R}\backslash\left\{\mathrm{3}\right\} \\ $$$$\left(\mathrm{2}\right)\:{a}\in\mathbb{C}\wedge{b}\in\mathbb{C};\:\sqrt[{\mathrm{5}}]{{x}}=\mid{x}\mid\mathrm{e}^{\mathrm{i}\frac{\mathrm{arg}\:{x}}{\mathrm{5}}} \\ $$$$\Rightarrow\:{a}\neq\mathrm{3}\:\Rightarrow\:{a}\in\mathbb{C}\backslash\left\{\mathrm{3}\right\} \\ $$
Commented by MathSh last updated on 28/Dec/20
$${Thanks}\:{sir} \\ $$