Question

If $a=\log _{2} 3\, b=\log _{2} 5, c=\log _{7} 2$, then $\log _{140} 63$ in terms of $a, b, c$ is

• A. $ \frac{2ac+1}{2a+abc+1} $
• B. $ \frac{2ac+1}{2a+c+a} $
• C. $ \frac{2ac+1}{2c+ab+a} $
• D. None of these

Answer 1

Answer: (D)

Now, $\log _{140} 63=\log _{2^{2} \times 5 \times 7}(3 \times 3 \times 7)$
$=\frac{\log _{2}(3 \times 3 \times 7)}{\log _{2}\left(2^{2} \times 5 \times 7\right)}$
$=\frac{\log _{2} 3+\log _{2} 3+\log _{2} 7}{2 \log _{2} 2+\log _{2} 5+\log _{2} 7}$
$=\frac{2 a+\frac{1}{c}}{2+b+\frac{1}{c}}=\frac{2 a c+1}{2 c+b c+1}$