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Q.

If $\log_{10} 2 = a, \log_{10} 3 = b then \log_{0 .72} (9 .6)$ in terms of a and b is equal to

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a

$\frac{2 a + 3 b - 1}{5 a + b - 2}$

b

$\frac{5 a + b - 1}{3 a + 2 b - 2}$

c

$\frac{3 a + b - 2}{2 a + 3 b - 1}$

d

$\frac{2 a + 5 b - 2}{3 a + b - 1}$

answer is B.

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Detailed Solution

$\begin{matrix} \log_{0 .72} (9 .6) = \log_{\frac{72}{100}} (\frac{96}{10}) = \frac{(\log_{10} 96 - \log_{10} 10)}{(\log_{10} 72 - \log_{10} 100)} \\ = \frac{\log_{10} (2^{5} \times 3) - 1}{\log_{10} (2^{3} \times 3^{2}) - 2} \\ = \frac{5 a + b - 1}{3 a + 2 b - 2} \end{matrix}$

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