Domain — Logarithmic Function

Domain — Logarithmic Function. Practice questions to deepen understanding of the domain of logarithmic functions. Online math practice with full solutions and step-by-step explanations.

Domain of logarithmic functions practice — argument > 0, log of rational, two logarithms, log of a square, and ln. Detailed explanations.

This practice covers: basic logarithm (> 0), logarithm with an expression, log of rational (sign table), two logarithms.

10 questions

Question 1
10.00 pts

📊 Basic logarithm:

What is the domain of \(f(x) = \log(x)\)?

Explanation:
📊 Basic logarithm

The function:

\(f(x) = \log(x)\)

⚠️ The golden rule:

Logarithm is defined only for positive numbers!

Condition:
\(x > 0\) ✓

(not ≥ 0 — strictly > 0!)

Domain: \((0, \infty)\)

⚠️ Why not 0?

\(\log(0)\) is undefined!

\(\log(x)\) asks: "to what power must 10 be raised to get x?"

No power satisfies \(10^? = 0\)

So 0 is forbidden! ✗

Why not negatives?

\(\log(-5)\) is undefined!

\(10^?\) is always positive!

No power satisfies \(10^? = -5\)

So negatives are forbidden! ✗

0not allowedallowed
Check:

• \(x = -1\): \(\log(-1)\) ✗
• \(x = 0\): \(\log(0)\) ✗
• \(x = 0.001\): \(\log(0.001) = -3\) ✓
• \(x = 1\): \(\log(1) = 0\) ✓
• \(x = 100\): \(\log(100) = 2\) ✓
Question 2
10.00 pts

📐 Logarithm with expression:

What is the domain of \(f(x) = \log(x-3)\)?

Explanation:
📐 Logarithm with expression

The function:

\(f(x) = \log(x-3)\)

Solution:

Condition:
The expression inside the log > 0

\(x - 3 > 0\)

\(x > 3\) ✓

Domain: \((3, \infty)\)

⚠️ Open bracket!

\(x = 3\) itself is forbidden!

because then: \(\log(0)\) ✗

Only \(x > 3\) is allowed

3not allowedallowed
Check:

• \(x = 2\): \(\log(-1)\) ✗
• \(x = 3\): \(\log(0)\) ✗
• \(x = 3.1\): \(\log(0.1)\) ✓
• \(x = 4\): \(\log(1) = 0\) ✓
• \(x = 13\): \(\log(10) = 1\) ✓

The rule:

\(\log(x-a)\)

Domain: \((a, \infty)\)

Everything above a (not including!)
Question 3
10.00 pts

📊÷ Log of a rational:

What is the domain of \(f(x) = \log\left(\frac{x+2}{x-1}\right)\)?

Explanation:
📊÷ Log of a rational expression

The function:

\(f(x) = \log\left(\frac{x+2}{x-1}\right)\)

Solution:

Condition:
The fraction must be strictly positive

\(\frac{x+2}{x-1} > 0\)

Also: \(x - 1 \neq 0\)

Critical values:
Numerator = 0: \(x = -2\)
Denominator = 0: \(x = 1\)

Sign table:

IntervalNumeratorDenominatorFraction
\(x < -2\)--+ ✓
\(x = -2\)0-30 — log(0) ✗
\(-2 < x < 1\)+-- ✗
\(x = 1\)30undef ✗
\(x > 1\)+++ ✓

Domain:
\((-\infty, -2) \cup (1, \infty)\) ✓

⚠️ Both endpoints are open!

• \(x = -2\): fraction = 0, but \(\log(0)\) undefined → excluded
• \(x = 1\): division by 0 → excluded

Logarithm requires strictly positive input, so fraction = 0 is not allowed!

Key point:

For logarithm: the argument must be > 0,
not ≥ 0!

So the numerator cannot equal 0 either.
Question 4
10.00 pts

📊📊 Two logarithms:

What is the domain of \(f(x) = \log(x+1) + \log(3-x)\)?

Explanation:
📊📊 Two logarithms

The function:

\(f(x) = \log(x+1) + \log(3-x)\)

Two conditions:

Condition 1 — first log:
\(x + 1 > 0\)
\(x > -1\) ✓

Condition 2 — second log:
\(3 - x > 0\)
\(3 > x\)
\(x < 3\) ✓

Intersection:
\(x > -1\) and \(x < 3\)

\(-1 < x < 3\) ✓

Domain: \((-1, 3)\)

Graphical explanation:

Condition 1: (→→→ from -1 upward
Condition 2: ←←←) up to 3

Intersection: (-1, 3) ✓

An open interval on both sides!

x>-1x<3Domain:-13
Check:

• \(x = -2\): \(\log(-1)\) ✗
• \(x = -1\): \(\log(0)\) ✗
• \(x = 0\): \(\log(1) + \log(3)\) ✓
• \(x = 3\): \(\log(0)\) ✗
• \(x = 4\): \(\log(-1)\) ✗
Question 5
10.00 pts

²📊 Log of a square:

What is the domain of \(f(x) = \log(x^2)\)?

Explanation:
²📊 Log of a square

The function:

\(f(x) = \log(x^2)\)

Solution:

Condition:
\(x^2 > 0\)

When is a square > 0?

Always! Except when x = 0!

\(x^2 > 0\) whenever \(x \neq 0\) ✓

Domain:
\(\mathbb{R} \setminus \{0\}\)

Or: \((-\infty, 0) \cup (0, \infty)\) ✓

Explanation:

• \(x = -5\): \(\log(25)\) ✓
• \(x = -1\): \(\log(1) = 0\) ✓
• \(x = 0\): \(\log(0)\) ✗
• \(x = 1\): \(\log(1) = 0\) ✓
• \(x = 5\): \(\log(25)\) ✓

Negatives allowed too! (the square makes them positive)

0allowedallowed
Useful identity:

\(\log(x^2) = 2\log|x|\)

So negatives work too!

(the absolute value converts them to positives)
Question 6
10.00 pts

÷📊 Log in denominator:

What is the domain of \(f(x) = \frac{1}{\log(x-2)}\)?

Explanation:
÷📊 Log in denominator

The function:

\(f(x) = \frac{1}{\log(x-2)}\)

⚠️ Two restrictions!

Restriction 1 — log defined:
\(x - 2 > 0\)
\(x > 2\) ✓

Restriction 2 — denominator ≠ 0:
\(\log(x-2) \neq 0\)

When is log = 0?
\(\log(y) = 0\) when \(y = 1\)

so: \(x - 2 \neq 1\)
\(x \neq 3\) ✓

Intersection:
\(x > 2\) and \(x \neq 3\)

Domain:
\((2, 3) \cup (3, \infty)\) ✓

⚠️ Forbidden points:

• \(x = 2\): \(\log(0)\) undefined ✗
• \(x = 3\): \(\frac{1}{0}\) division by zero ✗

Two forbidden points for different reasons!

23
Check:

• \(x = 2.5\): \(\frac{1}{\log(0.5)}\) ✓
• \(x = 3\): \(\frac{1}{0}\) ✗
• \(x = 4\): \(\frac{1}{\log(2)}\) ✓
Question 7
10.00 pts

🌿 Natural logarithm:

What is the domain of \(f(x) = \ln(5-2x)\)?

Explanation:
🌿 Natural logarithm

The function:

\(f(x) = \ln(5-2x)\)

Solution:

Condition:
\(5 - 2x > 0\)

\(5 > 2x\)

\(2x < 5\)

\(x < \frac{5}{2}\)

\(x < 2.5\) ✓

Domain:
\((-\infty, 2.5)\)

or: \((-\infty, \frac{5}{2})\) ✓

Reminder:

\(\ln\) = natural logarithm (base e)

Exactly the same rules as log!

The expression must be > 0 ✓

2.5allowednot allowed
Check:

• \(x = 0\): \(\ln(5)\) ✓
• \(x = 2\): \(\ln(1) = 0\) ✓
• \(x = 2.5\): \(\ln(0)\) ✗
• \(x = 3\): \(\ln(-1)\) ✗
Question 8
10.00 pts

√📊 Square root and log:

What is the domain of \(f(x) = \sqrt{x-1} + \log(x-2)\)?

Explanation:
√📊 Square root and log

The function:

\(f(x) = \sqrt{x-1} + \log(x-2)\)

Two conditions:

Condition 1 — square root:
\(x - 1 \geq 0\)
\(x \geq 1\) ✓

Condition 2 — log:
\(x - 2 > 0\)
\(x > 2\) ✓

Intersection:
\(x \geq 1\) and \(x > 2\)

The stricter condition: \(x > 2\) ✓

Domain: \((2, \infty)\)

Explanation:

If \(x > 2\):

• then \(x > 1\) (certainly!) ✓
• then the square root is defined ✓
• then the log is defined ✓

The stricter condition "absorbs" the other!

x≥1x>2Domain:2
Why not [1,∞)?

Because at \(x = 1.5\):

• square root: \(\sqrt{0.5}\) ✓
• log: \(\log(-0.5)\) ✗

The log is not defined!
Question 9
10.00 pts

💡 Tip:

What is the difference between square root and logarithm domains?

Explanation:
💡 Critical difference

Comparison table:

FunctionCondition0 allowed?
\(\sqrt{x}\)\(x \geq 0\)✅ yes
\(\log(x)\)\(x > 0\)❌ no

Examples:

Square root:
\(f(x) = \sqrt{x-3}\)
Domain: \([3, \infty)\) ✓
(3 allowed — closed bracket)

Logarithm:
\(f(x) = \log(x-3)\)
Domain: \((3, \infty)\) ✓
(3 forbidden — open bracket)

Why?

Square root of 0:
\(\sqrt{0} = 0\) ✓ defined!

Log of 0:
\(\log(0)\) undefined! ✗

No power satisfies \(10^? = 0\)

Quick rule:

Square root: [ closed bracket
Log: ( open bracket

Log is always stricter!

Combined:

\(f(x) = \sqrt{x-1} + \log(x-1)\)

Domain: \((1, \infty)\)

The log determines it! (stricter)
Question 10
10.00 pts

📚 Summary:

What is the single condition for a logarithm?

Explanation:
📚 Summary — Logarithm

The single rule:

The expression inside the log > 0

(not ≥ 0 — strictly > 0!)

Examples:

FunctionDomain
\(\log(x)\)\((0, \infty)\)
\(\log(x-a)\)\((a, \infty)\)
\(\log(a-x)\)\((-\infty, a)\)
\(\log(x^2)\)\(\mathbb{R} \setminus \{0\}\)
\(\frac{1}{\log(x)}\)\((0, 1) \cup (1, \infty)\)

With a rational expression:

\(\log\left(\frac{A(x)}{B(x)}\right)\)

Condition: \(\frac{A(x)}{B(x)} > 0\)

Use a sign table!

With a square root:

\(\sqrt{x} + \log(x)\)

• sqrt: \(x \geq 0\)
• log: \(x > 0\)

Domain: \((0, \infty)\)

The log is stricter!

⚠️ Remember:

• \(\log(0)\) undefined
• \(\log(-5)\) undefined
• Only \(\log(\text{positive})\) is defined!

\(\ln\) has the same rules exactly!