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Basics & Logarithms: Number Solutions Equation

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

Question
Number of solutions of the equation
log⁡3(3+x)+1log⁡(1+x2)33=1\log_3(3 + \sqrt{x}) + \frac{1}{\log_{(1+x^2)^3} 3} = 1
 is
Solution
Answer: 0
Step 1: Convert the second term. Using 1log⁡ba=log⁡ab\dfrac{1}{\log_b a} = \log_a b:
1log⁡(1+x2)33=log⁡3(1+x2)3=3log⁡3(1+x2)\frac{1}{\log_{(1+x^2)^3} 3} = \log_3(1+x^2)^3 = 3\log_3(1+x^2)
 Step 2: The equation becomes:
log⁡3(3+x)+3log⁡3(1+x2)=1  ⟹  log⁡3 ⁣[(3+x)(1+x2)3]=1\log_3(3+\sqrt{x}) + 3\log_3(1+x^2) = 1 \implies \log_3\!\left[(3+\sqrt{x})(1+x^2)^3\right] = 1
(3+x)(1+x2)3=3(3+\sqrt{x})(1+x^2)^3 = 3
 Step 3: The domain requires x≥0x \geq 0. For x≥0x \geq 0 (3+x)≥3(3 + \sqrt{x}) \geq 3 and (1+x2)3≥1(1+x^2)^3 \geq 1, so (3+x)(1+x2)3≥3(3+\sqrt{x})(1+x^2)^3 \geq 3. Equality holds only at x=0x = 0, but then the base of the second log is (1+0)3=1(1+0)^3 = 1, which is invalid (base equals 1). For x>0x > 0, the product strictly exceeds 3, so no solution exists. Answer: 0
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