Cho $\Large \log_{2}3=a, \log_{3}5=b, \log_{7}2=c$. Tính $\Large \log_

Cho $\Large \log_{2}3=a, \log_{3}5=b, \log_{7}2=c$. Tính $\Large \log_{140}63$ theo a, b, c

• A.

  A. $\Large \dfrac{2ac+1}{a+abc+2b}$

• B.

  B. $\Large \dfrac{2bc+1}{2c+abc+1}$

• C.

  C. $\Large \dfrac{2ac+1}{2c+abc+1}$

• D.

  D. $\Large \dfrac{3ab+1}{2a+abc+b}$

Chọn C

Ta có: $\Large \log_{140}63=\log_{2^2.5.7}(3^2.7)=\dfrac{2\log_{2}3+\log_{2}7}{2+\log_{2}5+\log_{2}7}$ và $\Large \log_{2}5=\log_{2}3.\log_{3}5=ab$

Vậy $\Large \log_{140}63=\dfrac{2a+\dfrac{1}{c}}{2+ab+\dfrac{1}{c}}=\dfrac{2ac+1}{2c+abc+1}$