If a = _ 2 3, b = _ 2 5 and c = _ 7 2 , then _ 140 63 in terms of...

We have,&

$\log_{140}63 = \log_{{2}^{2} \times 5 \times 7} ( 3 \times 3 \times 7 )$

$= \dfrac{\log_{2} ( 3 \times 3 \times 7 )}{\log_{2}\left({2}^{2} \times 5 \times 7\right)} = \dfrac{\log_{2}3 + \log_{2}3 + \log_{2}7}{2\log_{2}2 + \log_{2}5 + \log_{2}7}$

$= \dfrac{2a + \dfrac{1}{c}}{2 + b + \dfrac{1}{c}} = \dfrac{2ac + 1}{2c + bc + 1}$