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Question Number 205324 by hardmath last updated on 16/Mar/24

$$\mathrm{Compare}:$$$${37}^{37}\mathrm{and}{36}^{38}$$

Answered by nikif99 last updated on 16/Mar/24

$${37}^{37}\lessgtr {36}^{38}\Rightarrow$$$${37}^{36}\times 37\lessgtr {36}^{36}\times {36}^2\Rightarrow$$$$\frac{{37}^{36}}{{36}^{36}}\lessgtr \frac{{36}^2}{37}\Rightarrow$$$${\left(\frac{37}{36}\right)}^{36}\lessgtr \frac{{36}^2}{37}\Rightarrow$$$${(1+\frac{1}{36})}^{36}\lessgtr \frac{{36}^2}{37}\Rightarrow$$$$e<\frac{{36}^2}{37}$$

Answered by Frix last updated on 16/Mar/24

$$n\in \mathbb{N}$$$$1⩽n⩽4:n^n>{(n-1)}^{n+1}$$$$n⩾5:n^n<{(n-1)}^{n+1}$$

Answered by Berbere last updated on 17/Mar/24

$$\iff 37ln\left(37\right)and38ln\left(36\right)$$$$x=36;claim$$$$(x+1)ln(x+1)⩽(x+2)ln\left(x\right)$$$$\iff (x+1)(ln\left(x\right)+ln(1+\frac{1}{x}))⩽(x+2)ln\left(x\right)$$$$\iff (x+1)ln(1+\frac{1}{x})⩽ln\left(x\right)$$$$ln(1+a)={\int }_0^a\frac{1}{t+1}dt⩽{\int }_0^{a}dt=a$$$$\Rightarrow ln(1+\frac{1}{x})⩽\frac{1}{x}\Rightarrow (x+1)ln(1+\frac{1}{x})⩽(x+1).\frac{1}{x}=1+\frac{1}{x}⩽2$$$$\mathrm{\forall }x⩾1\Rightarrow 1+\frac{1}{x}⩽ln\left(x\right);\mathrm{\forall }x⩾e^2;e^2<36$$$$\Rightarrow {37}^{37}⩽{36}^{38}$$$$$$

Commented by hardmath last updated on 18/Mar/24

$$\mathrm{cool}\mathrm{dear}\mathrm{professors}\mathrm{thank}\mathrm{you}$$

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