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The value of \(log_a\bigg(\frac{a}{b}\bigg)+log_b\bigg(\frac{b}{a}\bigg)\), for \(1<a≤b\) cannot be equal to

Updated On: Jan 15, 2026
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The Correct Option is B

Solution and Explanation

Step 1: Expand and Simplify

Applying logarithmic identities:

\[ \log_a\left(\frac{a}{b}\right) = \log_a(a) - \log_a(b) = 1 - \log_a(b) \] \[ \log_b\left(\frac{b}{a}\right) = \log_b(b) - \log_b(a) = 1 - \log_b(a) \]

Step 2: Combine Expressions

\[ \log_a\left(\frac{a}{b}\right) + \log_b\left(\frac{b}{a}\right) = (1 - \log_a(b)) + (1 - \log_b(a)) \] \[ = 2 - (\log_a(b) + \log_b(a)) \]

Step 3: Substitute Variable

Let \( \log_a(b) = x \). By the change of base formula:

\[ \log_b(a) = \frac{1}{x} \] The expression becomes: \[ 2 - \left(x + \frac{1}{x} \right) \]

Step 4: Analyze the Function

Consider the function: \[ f(x) = x + \frac{1}{x} \] By the AM–GM Inequality, for \( x>0 \): \[ x + \frac{1}{x} \geq 2 \] Therefore, the original expression satisfies: \[ 2 - \left(x + \frac{1}{x} \right) \leq 2 - 2 \] \[ \Rightarrow 2 - \left(x + \frac{1}{x} \right) \leq 0 \] The maximum value of the expression is \( 2 - 2 = 0 \).

Final Conclusion

Since the expression is always less than or equal to 0, it can never be equal to 1.

✅ Final Answer:

\[ \boxed{1 \text{ is not a possible value}} \]

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