The question is from Log and Exponents. This question is about logarithm - inequalities. We need to find out the base value of the log. With every extra hour you log in for this topic, it becomes exponentially simpler. CAT Logarithms and Exponents is a favorite in CAT Exam, and appears more often than expected in the CAT Quantitative Aptitude section in the CAT Exam

Question 1: If log_{2}X + log_{4}X = log_{0.25}√6 and x > 0, then x is

- 6
^{-1/6} - 6
^{1/6} - 3
^{-1/3} - 6
^{1/3}

6

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log_{2}x + log_{4}x = log_{0.25}√6

We can rewrite the equation as:

log_{2}x + \\frac{log_2{x}}{log_2{4}}\\) = log_{0.25}√6

log_{2}x * \\frac{3}{2}\\) = log_{0.25}√6

=> log_{2}x * 3 = 2log_{0.25}√6

=> log_{2}x^{3} = -log_{4}6

=> log_{2}x^{3} = \\frac{-log_2{6}}{log_2{4}}\\)

=> log_{2}x^{3} = \\frac{-log_2{6}}{6}\\)

=> 2log_{2}x^{3} = -log_{2}6

2log_{2}x^{3} + log_{2}6 = 0

log_{2}6X^{6} = 0

6x^{6} = 1

x^{6} = \\frac{1}{6}\\)

x =

The question is **"If log _{2}X + log_{4}X = log_{0.25} \\sqrt {6}\\) and x > 0, then x is"**

Choice A is the correct answer.

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