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, ${\log }_{140}63={\log }_{22\times 5\times 7}(3\times 3\times 7)$
$<br/>=\frac{{\log }_2(3\times 3\times 7)}{{\log }_2\left(2^2\times 5\times 7\right)}=\frac{{\log }_{2}3+{\log }_{2}3+{\log }_{2}7}{2{\log }_{2}2+{\log }_{2}5+{\log }_{2}7}<br/>$
$=\frac{2a+\frac{1}{c}}{2+b+\frac{1}{c}}=\frac{2ac+1}{2c+bc+1}$