Question Number 97390 by shaxzod last updated on 07/Jun/20
$${compare}\:{log}_{\mathrm{2}} \mathrm{3}\:{with}\:{log}_{\mathrm{3}} \mathrm{4} \\ $$
Answered by mr W last updated on 07/Jun/20
$$\mathrm{9}>\mathrm{8} \\ $$$$\mathrm{3}^{\mathrm{2}} >\mathrm{2}^{\mathrm{3}} \\ $$$$\mathrm{2}\:\mathrm{log}_{\mathrm{2}} \:\mathrm{3}>\mathrm{3} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{2}} \:\mathrm{3}>\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{6}} \\ $$$$ \\ $$$$\mathrm{8}<\mathrm{9} \\ $$$$\mathrm{2}^{\mathrm{3}} =\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} <\mathrm{3}^{\mathrm{2}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{log}_{\mathrm{3}} \:\mathrm{4}<\mathrm{2} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{3}} \:\mathrm{4}<\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{8}}{\mathrm{6}}<\frac{\mathrm{9}}{\mathrm{6}}=\frac{\mathrm{3}}{\mathrm{2}}<\mathrm{log}_{\mathrm{2}} \:\mathrm{3} \\ $$$${i}.{e}.\:\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\:>\:\mathrm{log}_{\mathrm{3}} \:\mathrm{4} \\ $$