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Question Number 195101 by mathlove last updated on 24/Jul/23

$$\sqrt{ln2}\stackrel{?}{>}ln2$$

Answered by Frix last updated on 24/Jul/23

$${\mathrm{e}}^x=2$$$$\mathrm{e}>2\Rightarrow x<1$$$$x=\ln 2<1$$$$x<1\Rightarrow y=\frac{1}{x}>1;y=\frac{1}{x}\iff x=\frac{1}{y}\iff \sqrt{x}=\frac{1}{\sqrt{y}}$$$$y>1\Rightarrow \sqrt{y}<y\iff \frac{1}{\sqrt{y}}>\frac{1}{y}\stackrel{y=\frac{1}{x}}{\Rightarrow }\sqrt{x}>x\stackrel{x=\ln 2}{\Rightarrow }$$$$\sqrt{\ln 2}>\ln 2$$

Commented by mathlove last updated on 24/Jul/23

$$thanks$$

Commented by Frix last updated on 24/Jul/23

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Answered by mr W last updated on 25/Jul/23

$$for0<a<1,wehave$$$$a^2=a\times a<1\times a=a$$$$\Rightarrow a<\sqrt{a}$$$$2<e$$$$\Rightarrow \ln 2<\ln e=1$$$$\Rightarrow \ln 2<\sqrt{\ln 2}$$

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