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

Answer 1

Answer: (D)

Now, $\log _{140} 63=\log _{22 \times 5 \times 7}(3 \times 3 \times 7)$
$
=\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}
$
$=\frac{2 a +\frac{1}{ c }}{2+ b +\frac{1}{ c }}=\frac{2 ac +1}{2 c + bc +1}$