A step-by-step tutorial, with detailed solutions, on how to find the domain of real-valued logarithmic functions is presented. You can also use our Step-by-Step Calculator to Find Domain of a Function.
Definition of the Domain of a Function: For a function \( f \) defined by an expression with variable \( x \), the implied domain of \( f \) is the set of all real numbers variable \( x \) can take such that the expression defining the function is real. The domain can also be given explicitly.
Examples on How to Find the Domain of Logarithmic Functions
Example 1
Find the domain of function \( f \) defined by:
\[ f(x) = \log_3(x - 1) \]View Solution
\( f(x) \) can take real values if the argument of \( \log_3(x - 1) \), which is \( x - 1 \), is positive. Set up the condition on the argument:
\[ x - 1 > 0 \]Solve the inequality for \( x \) to obtain the domain:
\[ x > 1 \quad \text{or in interval form} \quad (1, +\infty) \]Example 2
Find the domain of function \( f \) defined by:
\[ f(x) = \log_2(x^2 + 5) \]View Solution
The argument of \( \log_2(x^2 + 5) \), which is \( x^2 + 5 \), is always greater than zero for any real number \( x \) because \( x^2 \geq 0 \). Therefore, the domain of the given function is:
\[ (-\infty, +\infty) \]Example 3
Find the domain of function \( f \) defined by:
\[ f(x) = \ln(9 - x^2) \]View Solution
For \( \ln(9 - x^2) \) to be real, the argument \( 9 - x^2 \) must be positive:
\[ 9 - x^2 > 0 \]Solving this inequality yields the interval:
\[ (-3, 3) \]Thus, the domain of the given function is \( (-3, 3) \).
Example 4
Find the domain of function \( f \) defined by:
\[ f(x) = \log_4|x - 3| \]View Solution
The domain of this function is the set of all values of \( x \) such that \( |x - 3| > 0 \). The expression \( |x - 3| \) is positive for all real values except for \( x = 3 \), which makes it zero. Hence, the domain is the set of all real numbers except 3:
\[ (-\infty, 3) \cup (3, +\infty) \]Or in inequality form:
\[ x < 3 \quad \text{or} \quad x > 3 \]Example 5
Find the domain of function \( f \) defined by:
\[ f(x) = \ln(2x^2 - 3x - 5) \]View Solution
The domain requires the argument to be strictly positive:
\[ 2x^2 - 3x - 5 > 0 \]Factor the expression on the left-hand side:
\[ (2x - 5)(x + 1) > 0 \]Solve the inequality to obtain the solution set:
\[ x < -1 \quad \text{or} \quad x > \frac{5}{2} \]The domain in interval form is:
\[ (-\infty, -1) \cup \left(\frac{5}{2}, +\infty\right) \]