I. $8^{{\log }_{2}3}={\left(2^3\right)}^{{\log }_{2}3}=2^{3{\log }_{2}3}=2^{{\log }_2{3}^3}=2^{{\log }_{2}27}=27$  

(zgodnie ze wzorem $a^{{\log }_{a}b}=b$ ).

II. ${\left({\log }_{5}1\right)}^3={\left({\log }_5{5}^0\right)}^3=0^3=0$  

(zgodnie ze wzorem ${\log }_{a}1=0$ ).

III. $\sqrt{{\log }_{3}81}={\left({\log }_3{3}^4\right)}^{\frac{1}{2}}=4^{\frac{1}{2}}={\left(2^2\right)}^{\frac{1}{2}}=2^{2\cdot \frac{1}{2}}=2^1=2$  

 

Odpowiedź: D