Question Number 86298 by 21042009 last updated on 28/Mar/20
$${let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{2} \\ $$$${x}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}=\pm\sqrt{\mathrm{2}} \\ $$$${then}\:{let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{4} \\ $$$${x}^{\mathrm{4}} =\mathrm{4} \\ $$$${x}=\pm\sqrt[{\mathrm{4}}]{\mathrm{4}}=\pm\sqrt{\mathrm{2}} \\ $$$${so}\:{we}\:{had}\:{prove}\:\mathrm{2}=\mathrm{4}\:{right}? \\ $$
Commented by john santu last updated on 28/Mar/20
$${hahaha}..{no} \\ $$
Commented by 20092104 last updated on 28/Mar/20
$${why} \\ $$
Commented by Kunal12588 last updated on 28/Mar/20
$$\mathrm{1}^{\mathrm{0}} =\mathrm{1},\:\mathrm{2}^{\mathrm{0}} =\mathrm{1} \\ $$$$\left.\mathrm{1}\neq\mathrm{2}\:\left({why}?\right)\::\right) \\ $$
Commented by john santu last updated on 28/Mar/20
$$\mathrm{10000000}^{\mathrm{0}} \:=\:\mathrm{e}^{\mathrm{0}} \\ $$$$\mathrm{10000000}\:=\:\mathrm{e}\: \\ $$$$\mathrm{why}? \\ $$
Commented by 20092104 last updated on 29/Mar/20
![domain of x^x^x^⋰ is [(1/e^e ), e^(1/e) ]](Q86536.png)
$${domain}\:{of}\:{x}^{{x}^{{x}^{\iddots} } } \:{is}\:\left[\frac{\mathrm{1}}{{e}^{{e}} },\:{e}^{\frac{\mathrm{1}}{{e}}} \right] \\ $$