Cho $\large \log_35=a$. Tính $\large \log_{729}\dfrac{1}{125}$ theo a

Cho $\large \log_35=a$. Tính $\large \log_{729}\dfrac{1}{125}$ theo a

• A. $\large -\dfrac{1}{2}a$
• B. $\large \dfrac{1}{2}a$
• C. $\large \dfrac{1}{2a}$
• D. $\large -\dfrac{1}{2a}$

Ta có: $\large \log_{729}\dfrac{1}{125}=\log_{3^6}5^{-3}=-\dfrac{1}{2}\log_35=-\dfrac{1}{2}a$

Vậy: $\large \log_{729}\dfrac{1}{125}=-\dfrac{1}{2}a$