Question Number 37411 by hari321 last updated on 12/Jun/18
$$\left(\sqrt{\left.\mathrm{2}\right)^{\sqrt{\mathrm{2}}} }\right. \\ $$
Answered by MJS last updated on 12/Jun/18
$$\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{2}}} =\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\sqrt{\mathrm{2}}} =\mathrm{2}^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} =\mathrm{2}^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \approx\mathrm{1}.\mathrm{632527} \\ $$
Commented by MrW3 last updated on 13/Jun/18
$${but}\:\mathrm{1}.\mathrm{632527}^{\mathrm{1}.\mathrm{632527}} =\mathrm{2}.\mathrm{2258726}\neq\mathrm{2} \\ $$$${hence}\:\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{2}}} \:{is}\:{not}\:{a}\:{solution}\:{of} \\ $$$${x}^{{x}} =\mathrm{2}. \\ $$
Commented by MJS last updated on 16/Jun/18
$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{think}\:\mathrm{or}\:\mathrm{say}\:\mathrm{it}\:\mathrm{was}… \\ $$
Commented by MrW3 last updated on 18/Jun/18
$${sorry},\:{I}\:{misunderstood}\:{you}. \\ $$