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Question Number 172323 by Giantyusuf last updated on 25/Jun/22

$$whichnumberisgreater$$$${22}^{55}\mathit{a}\mathit{n}\mathit{d}{55}^{22}???$$

Answered by nurtani last updated on 25/Jun/22

$${22}^{55}>{55}^{22}$$

Answered by floor(10²Eta[1]) last updated on 25/Jun/22

$$2^{55}.{11}^{55}>5^{22}.{11}^{22}$$$$2^{55}.{11}^{33}>5^{22}$$$$\mathrm{clearly}\mathrm{true}\mathrm{since}{11}^{33}>5^{33}>5^{22}$$$$\Rightarrow {22}^{55}>{55}^{22}$$

Answered by Gazella thomsonii last updated on 25/Jun/22

$$$$$$\mathrm{A}={22}^{55}\mathrm{B}={55}^{22}$$$$\mathrm{lnA}=55\ln \left(22\right)\ln \left(\mathrm{B}\right)=22\ln \left(55\right)$$$$\ln \left(22\right)\approx 3.091\ln \left(55\right)\approx 4.007$$$$\mathrm{lnA}=165.\mathrm{xxx}\mathrm{lnB}=88.\mathrm{xxx}$$$$\therefore \mathrm{B}<\mathrm{A}$$

Answered by MJS_new last updated on 25/Jun/22

$$f\left(x\right)=x^{\frac{1}{x}}\mathrm{has}\mathrm{its}\mathrm{maximum}\mathrm{at}x=\mathrm{e}\Rightarrow$$$$\mathrm{\forall }a,b\mid \mathrm{e}<a<b:a^{\frac{1}{a}}>b^{\frac{1}{b}}$$$$\iff$$$$\frac{\ln a}{a}>\frac{\ln b}{b}$$$$\iff$$$$b\ln a>a\ln b$$$$\iff$$$$a^b>b^a$$$$$$$$\mathrm{for}a,b\in \mathbb{N}\wedge a,b>1\mathrm{the}\mathrm{only}\mathrm{exceptions}\mathrm{are}$$$$2^3<3^2$$$$2^4=4^2$$

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