Mamy obliczyć:

${\log }_{\sqrt{3}}27+{\log }_{\sqrt{2}}8=\dots$

Obliczmy każdy z logarytmów z osobna:

${\log }_{\sqrt{3}}27={\log }_{\sqrt{3}}{3}^3=3{\log }_{\sqrt{3}}3=3\cdot 2=6$, bo ${\left(\sqrt{3}\right)}^2=3$

${\log }_{\sqrt{2}}8={\log }_{\sqrt{2}}{2}^3=3{\log }_{\sqrt{2}}2=3\cdot 2=6$, bo ${\left(\sqrt{2}\right)}^2=2$

Zatem:

$\dots =6+6=12$