Question Number 11005 by ajfour last updated on 06/Mar/17
$${prove}\:{that}\:\frac{\mathrm{9}}{\mathrm{4}}\:<\:\left(\mathrm{log}\:_{\mathrm{2}} \mathrm{3}\right)^{\mathrm{2}} \:<\:\frac{\mathrm{25}}{\mathrm{9}}\:\:. \\ $$
Answered by mrW1 last updated on 06/Mar/17
$$\mathrm{2log}_{\mathrm{2}} \:\mathrm{3}=\mathrm{log}_{\mathrm{2}} \:\mathrm{3}^{\mathrm{2}} =\mathrm{log}_{\mathrm{2}} \:\mathrm{9}>\mathrm{log}_{\mathrm{2}} \:\mathrm{8}=\mathrm{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{3}} =\mathrm{3} \\ $$ $$\Rightarrow\mathrm{log}_{\mathrm{2}} \:\mathrm{3}>\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\Rightarrow\left(\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\right)^{\mathrm{2}} >\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$ $$ \\ $$ $$\mathrm{3log}_{\mathrm{2}} \:\mathrm{3}=\mathrm{log}_{\mathrm{2}} \:\mathrm{3}^{\mathrm{3}} =\mathrm{log}_{\mathrm{2}} \:\mathrm{27}<\mathrm{log}_{\mathrm{2}} \:\mathrm{32}=\mathrm{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{5}} =\mathrm{5} \\ $$ $$\Rightarrow\mathrm{log}_{\mathrm{2}} \:\mathrm{3}<\frac{\mathrm{5}}{\mathrm{3}} \\ $$ $$\Rightarrow\left(\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\right)^{\mathrm{2}} <\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{9}} \\ $$ $$ \\ $$ $$\Rightarrow\frac{\mathrm{9}}{\mathrm{4}}<\left(\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\right)^{\mathrm{2}} <\frac{\mathrm{25}}{\mathrm{9}} \\ $$
Commented byajfour last updated on 07/Mar/17
$${Thanks},\:{so}\:{easy}\:{for}\:{you}! \\ $$