Plecam de la inegalitatea evidenta: [tex] \sqrt[]{8}\ \textless \ 3\ \textless \ 8,o,logaritmam,in,baza,2: log_{2} \sqrt{8}\ \textless \ log_{2}3\ \textless \ log_{2}8,sau,[/tex][tex] log_{2} 2^{ \frac{3}{2} }\ \textless \ log_{2}3\ \textless \ log_{2}2^3,deci \frac{3}{2}\ \textless \ log_{2}3\ \textless \ 3[/tex]. se tine cont cafunctia logaritmica in baza supraunitara este strict crescatoare.
