This is a question from Logarithms. Logarithms is a favorite in **CAT Exam** and appears more often than expected in the ** CAT Quantitative Aptitude section** in the CAT Exam. With every extra hour you log in for this topic, it becomes exponentially simpler.

Question 23 : The value of log_{0.008}√5 + log_{√3}81 – 7 is equal to :

- \\frac{1}{3})
- \\frac{2}{3})
- \\frac{5}{6})
- \\frac{7}{6})

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Here we have to find the value of log_{0.008} √5 + log_{√3} 81 – 7.

⟹ log_{0.008} √5 + log_{√3} 81 – 7

log_{0.008} √5 can be written in the terms of five and log_{√3} 81 can be written in the terms of 3.

where √5 = 5\^\frac{1}{2}) ⟹ 0.008 = \\frac{2^3}{10^3}) = 5^{-3}

⟹ log_{0.008} √5 + log_{√3} 81 – 7

⟹ \\frac{log_5 √5}{log_5 0.008}) + \\frac{log_3 81}{log_3 √3}) - 7

⟹ \\frac{\frac{-1}{2}}{-3}) + \\frac{4}{\frac{1}{2}}) - 7

⟹ \\frac{-1}{6}) + 1 = \\frac{5}{6})

Hence the value of log_{0.008}√5 + log_{√3}81 – 7 = \\frac{5}{6})

The question is **"The value of log _{0.008}√5 + log_{√3}81 – 7 is equal to :" **

Choice C is the correct answer

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