Question Number 133996 by mr W last updated on 26/Feb/21
$${show}\:{that}\:\sqrt{\mathrm{2}}<\mathrm{log}_{\mathrm{2}} \:\mathrm{3}<\sqrt{\mathrm{3}} \\ $$
Answered by som(math1967) last updated on 26/Feb/21
$$\:\:\mathrm{2}^{\sqrt{\mathrm{2}}} <\mathrm{3}<\mathrm{2}^{\sqrt{\mathrm{3}}} \\ $$ $${log}_{\mathrm{2}} \mathrm{2}^{\sqrt{\mathrm{2}}} <{log}_{\mathrm{2}} \mathrm{3}<{log}_{\mathrm{2}} \mathrm{2}^{\sqrt{\mathrm{3}}} \\ $$ $$\sqrt{\mathrm{2}}<{log}_{\mathrm{2}} \mathrm{3}<\sqrt{\mathrm{3}} \\ $$
Commented bysom(math1967) last updated on 26/Feb/21
$${Is}\:{my}\:{way}\:{correct}\:{sir}? \\ $$
Commented bymr W last updated on 26/Feb/21
$${we}\:{may}\:{not}\:{use}\:{calculator}.\:{how}\:{is}\:{it} \\ $$ $${obviour}\:{that}\:\mathrm{2}^{\sqrt{\mathrm{2}}} <\mathrm{3}<\mathrm{2}^{\sqrt{\mathrm{3}}} \:? \\ $$
Answered by mr W last updated on 26/Feb/21
$$\mathrm{9}>\mathrm{8} \\ $$ $$\mathrm{3}^{\mathrm{2}} >\mathrm{2}^{\mathrm{3}} \\ $$ $$\mathrm{2}\:\mathrm{log}_{\mathrm{2}} \:\mathrm{3}>\mathrm{3} \\ $$ $$\mathrm{log}_{\mathrm{2}} \:\mathrm{3}>\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\left(\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\right)^{\mathrm{2}} >\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}}>\frac{\mathrm{8}}{\mathrm{4}}=\mathrm{2} \\ $$ $$\Rightarrow\mathrm{log}_{\mathrm{2}} \:\mathrm{3}>\sqrt{\mathrm{2}} \\ $$ $$ \\ $$ $$\mathrm{27}<\mathrm{32} \\ $$ $$\mathrm{3}^{\mathrm{3}} <\mathrm{2}^{\mathrm{5}} \\ $$ $$\mathrm{3log}_{\mathrm{2}} \:\mathrm{3}<\mathrm{5} \\ $$ $$\mathrm{log}_{\mathrm{2}} \:\mathrm{3}<\frac{\mathrm{5}}{\mathrm{3}} \\ $$ $$\left(\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\right)^{\mathrm{2}} <\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{9}}<\frac{\mathrm{27}}{\mathrm{9}}=\mathrm{3} \\ $$ $$\Rightarrow\mathrm{log}_{\mathrm{2}} \:\mathrm{3}<\sqrt{\mathrm{3}} \\ $$
Commented bysom(math1967) last updated on 27/Feb/21
$${ok}\:{sir} \\ $$