CAT Quantitative Aptitude Questions | CAT Logarithms and Exponents

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 log₂X + log₄X = log_0.25√6 and x > 0, then x is

1. 6^-1/6
2. 6^1/6
3. 3^-1/3
4. 6^1/3

Explanatory Answer

Method of solving this CAT Question from Logs and Exponents: Logarithm with non-integer bases are slightly trickier. Have a go at this one!

log₂x + log₄x = log_0.25√6
We can rewrite the equation as:
log₂x + \\frac{log_2{x}}{log_2{4}}\\) = log_0.25√6
log₂x * \\frac{3}{2}\\) = log_0.25√6
=> log₂x * 3 = 2log_0.25√6
=> log₂x³ = -log₄6
=> log₂x³ = \\frac{-log_2{6}}{log_2{4}}\\)
=> log₂x³ = \\frac{-log_2{6}}{6}\\)

=> 2log₂x³ = -log₂6
2log₂x³ + log₂6 = 0
log₂6X⁶ = 0
6x⁶ = 1
x⁶ = \\frac{1}{6}\\)
x = $\sqrt[6]{1 \over 6}$

The question is "If log₂X + log₄X = log_0.25 \\sqrt {6}\\) and x > 0, then x is"

Hence, the answer is "6^-1/6".

Choice A is the correct answer.