The question is from Log and Exponents. We can solve this easily using log identities. We need to find out the value of x from the given equation. 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 7: log3x + logx3 = \\frac{17}{4}\\). Find the value of x.
 log3x + logx3 = \\frac{17}{4}\\)
 Let y = log3x
 We know that logx3 = \\frac{1}{log_3{x}}\\)
 Hence logx3 = \\frac{1}{y}\\)
 Thus the equation can be written as y + \\frac{1}{y}\\) = \\frac{17}{4}\\)
 4y2 + 4 = 17y
 4y2 + 4 - 17y = 0
 Solving the above equation we get y = 4 or \\frac{1}{4}\\)
 If y = 4
 log3x = 4
 Then x = 34
 If y = \\frac{1}{4}\\)
 log3x = \\frac{1}{4}\\)
 Then x = 31/4
The question is "Find the value of x."
Choice C is the correct answer.
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