Question:
The domain of the real valued function
\[
f(x)=\frac{\sqrt{\log_{0.5}(x-3)}}{\sqrt{x-1}}
\]
is
The domain of the real valued function
\[
f(x)=\frac{\sqrt{\log_{0.5}(x-3)}}{\sqrt{x-1}}
\]
is
Show Hint
For logarithms with base \(0\lt a\lt 1\), the inequality reverses in nature. Also, when a square root is in the denominator, its inside expression must be strictly positive.
Updated On: Jun 26, 2026
- \((3,4]\)
- \([4,\infty)\)
- \((1,\infty)\)
- \((1,3)\)
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The Correct Option is A
Solution and Explanation
Step 1: Condition for the logarithm to exist.
For \[ \log_{0.5}(x-3) \] to be defined, we need \[ x-3\gt 0 \] Therefore, \[ x\gt 3 \]
Step 2: Condition for the numerator square root to be real.
Since \[ \sqrt{\log_{0.5}(x-3)} \] is real, we require \[ \log_{0.5}(x-3)\geq 0 \] Here, the base is \[ 0.5=\frac{1}{2} \] which lies between \(0\) and \(1\).
For a logarithm with base between \(0\) and \(1\), \[ \log_{0.5}(x-3)\geq 0 \] implies \[ 0\lt x-3\leq 1 \] So, \[ 3\lt x\leq 4 \]
Step 3: Condition for the denominator square root.
The denominator is \[ \sqrt{x-1} \] Since it is in the denominator, we need \[ \sqrt{x-1}\neq 0 \] and \[ x-1\gt 0 \] Thus, \[ x\gt 1 \]
Step 4: Find the common domain.
From the numerator condition, \[ 3\lt x\leq 4 \] From the denominator condition, \[ x\gt 1 \] The common interval is \[ (3,4] \]
Step 5: Final conclusion.
Therefore, the domain is \[ \boxed{(3,4]} \]
For \[ \log_{0.5}(x-3) \] to be defined, we need \[ x-3\gt 0 \] Therefore, \[ x\gt 3 \]
Step 2: Condition for the numerator square root to be real.
Since \[ \sqrt{\log_{0.5}(x-3)} \] is real, we require \[ \log_{0.5}(x-3)\geq 0 \] Here, the base is \[ 0.5=\frac{1}{2} \] which lies between \(0\) and \(1\).
For a logarithm with base between \(0\) and \(1\), \[ \log_{0.5}(x-3)\geq 0 \] implies \[ 0\lt x-3\leq 1 \] So, \[ 3\lt x\leq 4 \]
Step 3: Condition for the denominator square root.
The denominator is \[ \sqrt{x-1} \] Since it is in the denominator, we need \[ \sqrt{x-1}\neq 0 \] and \[ x-1\gt 0 \] Thus, \[ x\gt 1 \]
Step 4: Find the common domain.
From the numerator condition, \[ 3\lt x\leq 4 \] From the denominator condition, \[ x\gt 1 \] The common interval is \[ (3,4] \]
Step 5: Final conclusion.
Therefore, the domain is \[ \boxed{(3,4]} \]
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