A step by step tutorial, with detailed solutions, on how to find the domain of real valued logarithmic functions is presented. Problems matched to the exercises with solutions at the bottom of the page are also presented. also a Step by Step Calculator to Find Domain of a Function is included.

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.

\[ f(x) = \log_3(x - 1) \]

\( f(x) \) can take real values if the argument of \( \log_3(x - 1) \) which is \( x - 1 \) is positive. Hence the condition on the argument

\( x - 1 > 0 \)

Solve the above inequality for \( x \) to obtain the domain: \( x > 1 \) or in interval form (1, \( \infty \))

The argument of \( \log_2(x^2 + 5) \) which is \( x^2 + 5 \) is always greater than zero and therefore positive. Hence the domain of the given function is given by the interval:

(-\( \infty \), \( +\infty \))

For \( \ln(9 - x^2) \) to be real, the argument of \( \ln(9 - x^2) \) which is \( 9 - x^2 \) must be positive. Hence the inequality

\( 9 - x^2 > 0 \)

The solution of the inequality \( 9 - x^2 > 0 \) is given by interval

\( (- 3, 3) \)

The domain of the given function is given by the interval \( (- 3, 3) \).

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 of the given function is the set of all real values except 3, which can be written in interval form as follows

(-\( \infty \), 3) \( \cup \) (3, \( +\infty \))

or in inequality form as follows

\( x \lt 3 \) or \( x > 3 \)

The domain of this function is the set of all values of \( x \) such that \( 2x^2 - 3x - 5 > 0 \). We need to solve the inequality

\( 2x^2 - 3x - 5 > 0 \)

Factor the expression on the left hand side of the inequality

\( (2x - 5)(x + 1) > 0 \) Solve the above inequality to obtain the solution set as follows:

\( x \lt -1 \) or \( x > \frac{5}{2} \)

The domain is given in inequality form as \( x \lt -1 \) or \( x > \frac{5}{2} \) and in interval form as follows:

(-\( \infty \), -1) \( \cup \) (\( \frac{5}{2} \), \( +\infty \))

- (-\( \infty \), 3)
- (-\( \infty \), \( +\infty \))
- (- 4, 4)
- (- \( \infty \), - 6) \( \cup \) (- 6, \( +\infty \))
- (- \( \infty \), -\(\frac{7}{3}\)) \( \cup \) (- 1, \( +\infty \))

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