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Basics & Logarithms: Real Satisfying Equation

JEE Maths question with a full step-by-step solution.

Question
Real xx satisfying the equation 9log⁡3(log⁡2x)=log⁡2x−(log⁡2x)2+19^{\log_3(\log_2 x)} = \log_2 x - (\log_2 x)^2 + 1 is
Solution
Answer: 2
Step 1: Simplify the left side. Using 9=329 = 3^2:
9log⁡3(log⁡2x)=32log⁡3(log⁡2x)=(log⁡2x)29^{\log_3(\log_2 x)} = 3^{2\log_3(\log_2 x)} = (\log_2 x)^2
 Step 2: Let t=log⁡2xt = \log_2 x. The equation becomes:
t2=t−t2+1  ⟹  2t2−t−1=0  ⟹  (2t+1)(t−1)=0t^2 = t - t^2 + 1 \implies 2t^2 - t - 1 = 0 \implies (2t + 1)(t - 1) = 0
 So t=−12t = -\dfrac{1}{2} or t=1t = 1. Step 3: Check validity. For the original equation, log⁡3(log⁡2x)\log_3(\log_2 x) must be defined, so log⁡2x>0\log_2 x > 0, meaning x>1x > 1. t=1⇒x=2>1t = 1 \Rightarrow x = 2 > 1. Valid. t=−12⇒x=2−1/2=12<1t = -\dfrac{1}{2} \Rightarrow x = 2^{-1/2} = \dfrac{1}{\sqrt{2}} < 1, so log⁡2x<0\log_2 x < 0, making log⁡3(log⁡2x)\log_3(\log_2 x) undefined. Rejected. Answer: 2
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