Domain of Definition
Domain of a Logarithmic Function
⭐ The Central Rule
The expression inside the logarithm must be strictly positive!
\(\log(\text{expression})\) → require: \(\text{expression} > 0\)
💡 Why?
The logarithm asks: "To what power must the base be raised to get the number?"
A positive base raised to any power gives only a positive number!
So there is no logarithm of 0 or a negative number.
⚠️ Note the difference:
Root: ≥ 0 (including zero)
Logarithm: > 0 (not including zero!)
📐 Basic Examples
Example 1: \(f(x) = \log(x)\)
Condition: \(x > 0\)
Domain: \((0, \infty)\)
Example 2: \(f(x) = \ln(x)\) (natural log)
Condition: \(x > 0\)
Domain: \((0, \infty)\)
Example 3: \(f(x) = \log(x - 3)\)
Condition: \(x - 3 > 0\) → \(x > 3\)
Domain: \((3, \infty)\)
Example 4: \(f(x) = \ln(5 - x)\)
Condition: \(5 - x > 0\) → \(x < 5\)
Domain: \((-\infty, 5)\)
Example 5: \(f(x) = \log(2x + 8)\)
Condition: \(2x + 8 > 0\) → \(x > -4\)
Domain: \((-4, \infty)\)
📈 Quadratic Expression Inside the Log
Example 6: \(f(x) = \log(x^2 - 4)\)
Condition: \(x^2 - 4 > 0\)
Factoring: \((x-2)(x+2) > 0\)
Parabola with a > 0 → positive "outside"
Domain: \(x < -2\) or \(x > 2\)
Example 7: \(f(x) = \ln(-x^2 + 6x - 5)\)
Condition: \(-x^2 + 6x - 5 > 0\)
Multiply by (−1): \(x^2 - 6x + 5 < 0\)
Factoring: \((x-1)(x-5) < 0\)
Parabola with a > 0 → negative "inside"
Domain: \(1 < x < 5\)
Example 8: \(f(x) = \log(x^2 + 1)\)
Condition: \(x^2 + 1 > 0\)
Check: \(x^2 + 1 \geq 1 > 0\) always!
Domain: ℝ (all real numbers)
🔮 Log Combined with a Root
Example 9: \(f(x) = \sqrt{\ln(x)}\)
Condition 1 (ln defined): \(x > 0\)
Condition 2 (root): \(\ln(x) \geq 0\) → \(x \geq 1\)
Intersection: \(x > 0\) and \(x \geq 1\)
Domain: \(x \geq 1\)
Example 10: \(f(x) = \sqrt{1 - \ln(x)}\)
Condition 1 (ln defined): \(x > 0\)
Condition 2 (root): \(1 - \ln(x) \geq 0\)
\(\ln(x) \leq 1\) → \(x \leq e\)
Intersection: \(x > 0\) and \(x \leq e\)
Domain: \(0 < x \leq e\)
Example 11: \(f(x) = \ln(\sqrt{x} - 1)\)
Condition 1 (root): \(x \geq 0\)
Condition 2 (log): \(\sqrt{x} - 1 > 0\)
\(\sqrt{x} > 1\) → \(x > 1\)
Intersection: \(x \geq 0\) and \(x > 1\)
Domain: \(x > 1\)
🔴 Log Combined with a Rational Function
Example 12: \(f(x) = \frac{1}{\ln(x)}\)
Condition 1 (ln defined): \(x > 0\)
Condition 2 (denominator): \(\ln(x) \neq 0\) → \(x \neq 1\)
Domain: \(x > 0\) and \(x \neq 1\)
Example 13: \(f(x) = \ln\left(\frac{x}{x-2}\right)\)
Condition: \(\frac{x}{x-2} > 0\)
The fraction is positive when numerator and denominator have the same sign:
Both positive: \(x > 0\) and \(x > 2\) → \(x > 2\)
Both negative: \(x < 0\) and \(x < 2\) → \(x < 0\)
Domain: \(x < 0\) or \(x > 2\)
Example 14: \(f(x) = \frac{\ln(x)}{x - 3}\)
Condition 1 (ln defined): \(x > 0\)
Condition 2 (denominator): \(x \neq 3\)
Domain: \(x > 0\) and \(x \neq 3\)
🔷 Complex Combinations
Example 15: \(f(x) = \frac{1}{\sqrt{\ln(x)}}\)
Condition 1 (ln defined): \(x > 0\)
Condition 2 (root in denominator): \(\ln(x) > 0\) → \(x > 1\)
Domain: \(x > 1\)
Example 16: \(f(x) = \ln(x) + \ln(4-x)\)
Condition 1: \(x > 0\)
Condition 2: \(4 - x > 0\) → \(x < 4\)
Intersection: \(x > 0\) and \(x < 4\)
Domain: \(0 < x < 4\)
Example 17: \(f(x) = \sqrt{x-1} + \ln(3-x)\)
Root condition: \(x \geq 1\)
Log condition: \(3 - x > 0\) → \(x < 3\)
Intersection: \(x \geq 1\) and \(x < 3\)
Domain: \(1 \leq x < 3\) or \([1, 3)\)
📝 Summary
Logarithmic function: inside the log must be > 0
Note: not ≥ 0 but > 0 (zero is excluded!)
With combinations — find the intersection of all conditions
ln(x) ≥ 0 ↔ x ≥ 1 | ln(x) ≤ 0 ↔ 0 < x ≤ 1