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Why log base 2 of (x^2 + 3) Never Equals a Negative Logarithm | JEE Advanced

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

Question

Number of solution(s) of the equation

\log_{2}\left(x^{2}+3\right) = \dfrac12\log_{1/3}\left(x+\dfrac1x\right)

x > 0

0

correct

1

2

Dinfinite

Solution

Step 1: Bound the left side from below. For every real

x

x^{2}+3 \ge 3 > 1 ,

and

\log_{2}

is increasing with

\log_{2}1 = 0

, so

\log_{2}\left(x^{2}+3\right) \ge \log_{2}3 > 1 > 0 .

The left side is always **positive**. Step 2: Bound the argument on the right. For

x > 0

, AM-GM gives

x + \frac1x \ge 2 .

Step 3: Bound the right side from above. The base

\dfrac13

is less than

1

, so

\log_{1/3}

is a **decreasing** function and

\log_{1/3}1 = 0

. Since its argument is at least

2 > 1

\log_{1/3}\left(x+\frac1x\right) \le \log_{1/3}2 < 0 ,

and halving keeps it negative. The right side is always **negative**. Step 4: Compare. A positive quantity can never equal a negative one, so the equation has no solution at all. Answer: (1).

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