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If $a = \log 2, b = \log 3, c = \log 7$ and $6^{x} = 7^{x + 4}$ then $x =$

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$\frac{4 b}{c - a - b}$

$\frac{4 a}{a + b - c}$

$\frac{4 b}{c + a - b}$

$\frac{4 c}{a + b - c}$

answer is B.

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Detailed Solution

$6^{x} = 7^{x + 4} x \log 6 = (x + 4) \log 7 x = \frac{4 \log 7}{\log 6 - \log 7} = \frac{4 \log 7}{\log 2 + \log 3 - \log 7} = \frac{4 c}{a + b - c}$

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