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

\frac{2 a c + 1}{2 c + b c + 1}

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\frac{2 a c + 1}{2 a + c + a}

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\frac{2 a c + 1}{2 c + a b + a}

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None of these

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Solution

The correct option is A $\frac{2 a c + 1}{2 c + b c + 1}$
Now,
${log}_{140} 63 = {log}_{2^{2} \times 5 \times 7} 3 \times 3 \times 7$
$= \frac{{log}_{2} 3 \times 3 \times 7}{2^{2} \times 5 \times 7} = \frac{{log}_{2} 3 + {log}_{2} 3 + {log}_{2} 7}{2 {log}_{2} 2 + {log}_{2} 5 + {log}_{2} 7}$
$= \frac{2 a + \frac{1}{c}}{2 + b + \frac{1}{c}} = \frac{2 a c + 1}{2 c + b c + 1}$

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