Question Number 195101 by mathlove last updated on 24/Jul/23
$$\sqrt{{ln}\mathrm{2}}\:\:\overset{?} {>}{ln}\mathrm{2} \\ $$
Answered by Frix last updated on 24/Jul/23
$$\mathrm{e}^{{x}} =\mathrm{2} \\ $$$$\mathrm{e}>\mathrm{2}\:\Rightarrow\:{x}<\mathrm{1} \\ $$$${x}=\mathrm{ln}\:\mathrm{2}\:<\mathrm{1} \\ $$$${x}<\mathrm{1}\:\Rightarrow\:{y}=\frac{\mathrm{1}}{{x}}>\mathrm{1};\:{y}=\frac{\mathrm{1}}{{x}}\:\Leftrightarrow\:{x}=\frac{\mathrm{1}}{{y}}\:\Leftrightarrow\:\sqrt{{x}}=\frac{\mathrm{1}}{\:\sqrt{{y}}} \\ $$$${y}>\mathrm{1}\:\Rightarrow\:\sqrt{{y}}<{y}\:\Leftrightarrow\:\frac{\mathrm{1}}{\:\sqrt{{y}}}>\frac{\mathrm{1}}{{y}}\:\overset{{y}=\frac{\mathrm{1}}{{x}}} {\Rightarrow}\:\sqrt{{x}}>{x}\:\overset{{x}=\mathrm{ln}\:\mathrm{2}} {\Rightarrow} \\ $$$$\sqrt{\mathrm{ln}\:\mathrm{2}}>\mathrm{ln}\:\mathrm{2} \\ $$
Commented by mathlove last updated on 24/Jul/23
$${thanks} \\ $$
Commented by Frix last updated on 24/Jul/23
��
Answered by mr W last updated on 25/Jul/23
$${for}\:\mathrm{0}<{a}<\mathrm{1},\:{we}\:{have} \\ $$$${a}^{\mathrm{2}} ={a}×{a}<\mathrm{1}×{a}={a} \\ $$$$\Rightarrow{a}<\sqrt{{a}} \\ $$$$\mathrm{2}<{e} \\ $$$$\Rightarrow\mathrm{ln}\:\mathrm{2}<\mathrm{ln}\:{e}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{ln}\:\mathrm{2}<\sqrt{\mathrm{ln}\:\mathrm{2}} \\ $$