Cho $\Large \log_{12}3=a$. Tính $\Large \log_{24}18$ theo $\Large a$ A

Cho $\Large \log_{12}3=a$. Tính  $\Large \log_{24}18$ theo $\Large a$

 

• A.

  A.  $\Large \dfrac{3a-1}{3-a}$

   
• B.

  B.  $\Large \dfrac{3a+1}{3-a}$

   
• C.

  C.  $\Large \dfrac{3a+1}{3+a}$

   
• D.

  D.  $\Large \dfrac{3a-1}{3+a}$

Chọn B

+) $\Large a=\log_{12}3=\dfrac{\log_{2}3}{\log_{2}12}=\dfrac{\log_{2}3}{\log_{2}(2^2.3)}$

$\Large =\dfrac{\log_{2}3}{\log_{2}2^2+\log_{2}3}=\dfrac{\log_{2}3}{2+\log_{2}3}$

$\Large \Rightarrow \log_{2}3=\dfrac{2a}{1-a}$

+) $\Large \log_{24}18=\dfrac{\log_{2}18}{\log_{2}24}=\dfrac{\log_{2}(2.3^2)}{\log_{2}(2^3.3)}=\dfrac{1+2\log_{2}3}{3+\log_{2}3}=\dfrac{1+2.\dfrac{2a}{1-a}}{3+\dfrac{2a}{1-a}}=\dfrac{3a+1}{3-a}$