Question:medium

The range of the function \(f(x) = \log_5(25 - x^2)\) is

Show Hint

Always find the range of the inner function first, then apply the logarithm. Log functions convert \((0,a]\) into \((-\infty, \log a]\).
Updated On: Jun 17, 2026
  • \([0,5]\)
  • \([0,2)\)
  • \((0,2)\)
  • None of these
Show Solution

The Correct Option is D

Solution and Explanation

To determine the range of the function \(f(x) = \log_5(25 - x^2)\), we need to analyze its domain and the output values it can take.

  1. The argument of the logarithm, \(25 - x^2\), must be greater than zero because the logarithm is only defined for positive arguments. Therefore, we set up the inequality: \(25 - x^2 > 0\) 
  2. Solving the inequality \(25 - x^2 > 0\) gives: \(25 > x^2\), which simplifies to: \(-5 < x < 5\)
  3. Within this range, \(x^2\) can vary from \(0\) (when \(x = 0\)) to a little less than \(25\) as \(x\) approaches \(5\) or \(-5\). Therefore, \(25 - x^2\) varies from \(25\) down to just above \(0\).
  4. Next, we consider \(f(x) = \log_5(25 - x^2)\). Since \(\log_5(y)\) (where \(y = 25-x^2\)) is a logarithmic function which is defined for \(y > 0\), we analyze its range:
    • When \(y = 25\), \(\log_5(25) = \log_5(5^2) = 2\).
    • As \(y\) approaches \(0\) (but remains positive, say \(y \to 0^{+}\)), \(\log_5(y) \to -\infty\).

Since none of the given options match the calculated range, the correct answer is indeed "None of these".

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