If \( \Large log_{8}x + log_{8}\frac{1}{6} = \frac{1}{3} \), then the value of x is
Correct Answer: Description for Correct answer:
\( \Large \log_{8}x + \log_{8}\frac{1}{6} = \frac{1}{3} \)
=> \( \Large \log_{8} (x \times \frac{1}{6}) = \frac{1}{3} \)
=> \( \Large \log_{8} (\frac{x}{6}) = \frac{1}{3} \)
=> \( \Large (8)^{\frac{1}{3}} = \frac{x}{6} \)
\( \Large [\because if \log_{x}y = x, \ then (a)^{x} = y] \)
=> \( \Large (2^{3})^{\frac{1}{3}} = \frac{x}{6} \)
=> \( \Large x = 12 \)
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