Question

The domain of the real valued function \( f(x)=\log_{x-1}(3x+1) \) is

• A. \( (1,\infty) \)
• B. \( \mathbb{R} \)
• C. \( (1,2)\cup(2,\infty) \)
• D. \( \mathbb{R} - \{2\} \)

Hint:

When finding the domain of a logarithmic function \( \log_{b} a \), always remember the core restrictions: \( a > 0 \), \( b > 0 \), and \( b \neq 1 \). Missing the condition \( b \neq 1 \) is a common mistake that leads to an incorrect choice.

Answer 1

The Correct Option is C

Concept: For a logarithmic function of the form \( \log_{b(x)} g(x) \) to be well-defined for real values, the following three conditions must be satisfied simultaneously:

• The argument must be strictly positive: \( g(x) > 0 \)

• The base must be strictly positive: \( b(x) > 0 \)

• The base cannot be equal to one: \( b(x) \neq 1 \)

Step 1: Condition for the argument \( (3x + 1) \).
The argument of the logarithm must be strictly greater than zero: \[ 3x + 1 > 0 \quad \Rightarrow \quad 3x > -1 \quad \Rightarrow \quad x > -\frac{1}{3} \quad \cdots (1) \]

Step 2: Condition for the base \( (x - 1) \).
The base of the logarithm must be strictly greater than zero and cannot be equal to one: \[ x - 1 > 0 \quad \Rightarrow \quad x > 1 \quad \cdots (2) \] \[ x - 1 \neq 1 \quad \Rightarrow \quad x \neq 2 \quad \cdots (3) \]

Step 3: Finding the intersection of all conditions.
Combining the inequalities from equations (1), (2), and (3):

• From \( x > -\frac{1}{3} \) and \( x > 1 \), the overlapping interval is \( x > 1 \), which can be written as \( (1, \infty) \).

• Excluding the value \( x = 2 \) from this interval gives the final domain.
Thus, the domain of the function is \( (1, 2) \cup (2, \infty) \).