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The value of $\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$ is equal to

  1. $1/3$
  2. $2/3$
  3. $5/6$
  4. $7/6$
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Given that, $\log_{0.008}{5^\frac{1}{2}} + \log_{3^\frac{1}{2}}{81}  – 7$

We know that, $81 = 3^{4} , 0.008 = \frac{8}{1000} = \frac{1}{125} = \frac{1}{5^{3}} = 5^{-3}$

Now, $\log_{0.008}{5^\frac{1}{2}} + \log_{3^\frac{1}{2}}{81}  – 7 = \log_{5^{-3}}{5^{\frac{1}{2}}} + \log_{3^\frac{1}{2}}{3^{4}} – 7$

$\qquad  = \frac{1}{-3} \cdot \frac{1}{2} \log_{5} 5 + 2 \cdot 4 \log_{3}3– 7 \quad [\because \log_{b}x^{a} = a \log_{b}x,  \log_{b^{a}}x = \frac{1}{a}\log_{b}x]$

$\qquad = \frac{-1}{6} + 8 – 7 \quad [\because \log_{a}a = 1]$

$\qquad = \frac{-1}{6} + 1 = \frac{-1+6}{6} = \frac{5}{6}.$

Correct Answer $: \text{C}$
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