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Basics & Logarithms — JEE Maths practice question

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

Question
Number of integral solutions of the equation log⁡x−3(log⁡2x2−2x+3(x2+2x))=0\log_{x-3}\left(\log_{2x^2-2x+3}(x^2+2x)\right) = 0 is:
A44
B22
C11
D00correct
Solution
Step 1: Removing the outer logarithm
log⁡2x2−2x+3(x2+2x)=(x−3)0=1  ⟹  x2+2x=2x2−2x+3\log_{2x^2-2x+3}(x^2+2x) = (x-3)^0 = 1 \implies x^2+2x = 2x^2-2x+3
 Step 2: Solve
x2−4x+3=0  ⟹  (x−3)(x−1)=0  ⟹  x=3 or x=1x^2-4x+3 = 0 \implies (x-3)(x-1) = 0 \implies x = 3 \text{ or } x = 1
 Step 3: Check the domain For x=3x = 3: outer base x−3=0x-3 = 0 (invalid). For x=1x = 1: outer base x−3=−2<0x-3 = -2 < 0 (invalid). Neither is admissible, so there is no solution. Answer: (4)
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