Question Number 176141 by Linton last updated on 13/Sep/22
$${which}\:{is}\:{greater}\:? \\ $$$$\mathrm{4}^{\sqrt{\mathrm{2}}} \:{or}\:\mathrm{3}^{\sqrt{\mathrm{3}}} \\ $$
Commented by Linton last updated on 13/Sep/22
$${without}\:{calculator}\:{show}\:{working} \\ $$
Answered by Frix last updated on 13/Sep/22
$$\mathrm{we}\:\mathrm{know}\:{x}>\mathrm{1}\wedge{y}>\mathrm{1}\wedge{z}>\mathrm{1}\wedge{x}\gtrless{y}\:\Rightarrow\:{x}^{{z}} \gtrless{y}^{{z}} \\ $$$$\mathrm{we}\:\mathrm{also}\:\mathrm{know}\:\mathrm{4}^{\sqrt{\mathrm{2}}} >\mathrm{1}\wedge\mathrm{3}^{\sqrt{\mathrm{3}}} >\mathrm{1}\wedge\sqrt{\mathrm{2}}>\mathrm{1} \\ $$$$\mathrm{4}^{\sqrt{\mathrm{2}}} \overset{?} {\gtrless}\mathrm{3}^{\sqrt{\mathrm{3}}} \\ $$$$\left(\mathrm{4}^{\sqrt{\mathrm{2}}} \right)^{\sqrt{\mathrm{2}}} \overset{?} {\gtrless}\left(\mathrm{3}^{\sqrt{\mathrm{3}}} \right)^{\sqrt{\mathrm{2}}} \\ $$$$\mathrm{16}\overset{?} {\gtrless}\mathrm{3}^{\sqrt{\mathrm{6}}} \\ $$$$\mathrm{we}\:\mathrm{know}\:\sqrt{\mathrm{6}}<\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{because}\:\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{4}}>\mathrm{6} \\ $$$$\mathrm{3}^{\frac{\mathrm{5}}{\mathrm{2}}} =\mathrm{9}\sqrt{\mathrm{3}} \\ $$$$\mathrm{we}\:\mathrm{know}\:\sqrt{\mathrm{3}}<\frac{\mathrm{7}}{\mathrm{4}}\:\mathrm{because}\:\left(\frac{\mathrm{7}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{49}}{\mathrm{16}}>\mathrm{3} \\ $$$$\mathrm{but}\:\mathrm{9}×\frac{\mathrm{7}}{\mathrm{4}}=\frac{\mathrm{63}}{\mathrm{4}}<\mathrm{16} \\ $$$$\Rightarrow \\ $$$$\mathrm{16}>\mathrm{9}×\frac{\mathrm{7}}{\mathrm{4}}>\mathrm{9}\sqrt{\mathrm{3}}>\mathrm{3}^{\sqrt{\mathrm{6}}} \\ $$$$\mathrm{16}>\mathrm{3}^{\sqrt{\mathrm{6}}} \\ $$$$\mathrm{4}^{\sqrt{\mathrm{2}}} >\mathrm{3}^{\sqrt{\mathrm{3}}} \\ $$
Commented by Linton last updated on 14/Sep/22
$${well}\:{done} \\ $$