Q. 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) $\frac{2ac + 1}{2c + abc + 1}$
B) $\frac{2ac + 1}{2a + c + a}$
C) $\frac{2ac + 1}{2c + ab + a}$
D) None of these
Explanation
(d)
$\log_{140}{63 = \log_{2^{2} \times 5 \times 7}{(3 \times 3 \times 7)}}$
$= \frac{\log_{2}{(3 \times 3 \times 7)}}{\log_{2}{(2^{2} \times 5 \times 7)}}$
$= \frac{2\log_{2}{3 + \log_{2}7}}{2\log_{2}{2 + \log_{2}{5 + \log_{2}7}}}$
$= \frac{2a + \frac{1}{c}}{2 + b + \frac{1}{c}} = \frac{2ac + 1}{2c + bc + 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.