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Question Number 183660 by mr W last updated on 28/Dec/22

$$whatislarger,$$$${2022}^{2022}or{2021}^{2023}?$$

Answered by Ar Brandon last updated on 28/Dec/22

$$1^1>0^2$$$$2^2>1^3$$$$3^3>2^4$$$$4^4>3^5$$$$5^5<4^6$$$$6^6<5^7$$$$x^x<{(x-1)}^{x+1}\mathrm{\forall }x\in \mathbb{N}\wedge x⩾5$$$$\Rightarrow {2022}^{2022}<{2021}^{2023}$$

Answered by manxsol last updated on 29/Dec/22

$${2022}^{2022}and2021\times {2021}^{2022}$$$$\frac{{2022}^{2022}}{{2021}^{2022}}and2021$$$${(1+\frac{1}{2021})}^{2022}={(1+\frac{1}{2021})}^{2021}\times (1+\frac{1}{2021})$$$$=e(1+\frac{1}{2021})=e+\frac{e}{2021}$$$$e+\frac{e}{2021}<2021$$$${2022}^{2022}⟨{2021}^{2023}$$$$theory$$$$li\underset{x\to \mathrm{\infty }}{m}{(1+\frac{1}{x})}^x=e=2.71...$$$$$$$$$$

Answered by mr W last updated on 28/Dec/22

$$A={2022}^{2022}$$$$B={2021}^{2023}$$$$\frac{A}{B}={\left(\frac{2022}{2021}\right)}^{2021}\times \left(\frac{2022}{2021}\right)\times \frac{1}{2021}$$$$={(1+\frac{1}{2021})}^{2021}\times \left(\frac{2022}{2021}\right)\times \frac{1}{2021}$$$$<e\times \left(\frac{2022}{2021}\right)\times \frac{1}{2021}$$$$<\frac{2022\times 3}{2021\times 2021}<1$$$$\Rightarrow A<B$$

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