Google
Câu hỏi: Cho $a,b,c>0,a\ne 1$ và ${{\log }_{a}}b=\sqrt{2022}$. Tính ${{\log }_{\sqrt[6]{a}}}\left( {{a}^{\dfrac{7}{4}}}.\sqrt[6]{b} \right).$
A. $42+\dfrac{\sqrt{2022}}{6}$.
B. $\dfrac{7}{4}+6\sqrt{2022}$.
C. $\dfrac{21}{2}+\sqrt{2022}$.
D. $\dfrac{2}{21}+\sqrt{2022}$.
Ta có: ${{\log }_{\sqrt[6]{a}}}\left( {{a}^{\dfrac{7}{4}}}.\sqrt[6]{b} \right)={{\log }_{\sqrt[6]{a}}}{{a}^{\dfrac{7}{4}}}+{{\log }_{\sqrt[6]{a}}}\sqrt[6]{b}=6.\dfrac{7}{4}+\sqrt{2022}=\dfrac{21}{2}+\sqrt{2022}.$
A. $42+\dfrac{\sqrt{2022}}{6}$.
B. $\dfrac{7}{4}+6\sqrt{2022}$.
C. $\dfrac{21}{2}+\sqrt{2022}$.
D. $\dfrac{2}{21}+\sqrt{2022}$.
Ta có: ${{\log }_{\sqrt[6]{a}}}\left( {{a}^{\dfrac{7}{4}}}.\sqrt[6]{b} \right)={{\log }_{\sqrt[6]{a}}}{{a}^{\dfrac{7}{4}}}+{{\log }_{\sqrt[6]{a}}}\sqrt[6]{b}=6.\dfrac{7}{4}+\sqrt{2022}=\dfrac{21}{2}+\sqrt{2022}.$
Đáp án C.