The domain of a logarithmic function is every x-value that keeps the argument (the expression inside the log) strictly positive. The range, for any standard logarithmic function, is all real numbers. That second part surprises many students because it never changes for basic transformations, but finding the domain requires real work depending on what’s inside the log.
The Core Rule for Domain
You can only take the logarithm of a positive number. Zero and negative numbers are not allowed. This means the expression inside the logarithm, called the argument, must be greater than zero. For the simplest case, f(x) = log(x), the domain is (0, ∞) in interval notation. The base of the logarithm, whether it’s 10, e, 2, or any other valid base, does not change this rule. Natural log, common log, and log base 2 all share the same domain restriction.
This single rule is the key to every domain problem you’ll encounter. No matter how complicated the expression inside the log gets, your job is always the same: set that expression greater than zero and solve the inequality.
Finding Domain With Linear Arguments
When the argument is a simple linear expression, set it greater than zero and solve for x. Here are a few examples showing how this works in practice.
For f(x) = log₂(x + 3), set x + 3 > 0 and solve to get x > −3. The domain is (−3, ∞). For f(x) = log₄(2x − 3), set 2x − 3 > 0 and solve to get x > 1.5. The domain is (1.5, ∞). Notice that in interval notation, you always use a parenthesis (not a bracket) on the boundary value, because the argument can get close to zero but never equal it.
For f(x) = 2log(3 − x) − 1, the argument is 3 − x. Setting 3 − x > 0 gives you x < 3, so the domain is (−∞, 3). When the variable is subtracted rather than added, the inequality flips direction, and the domain extends to the left instead of to the right.
Reflections That Reverse the Domain
A negative sign in front of x inside the log creates a reflection across the y-axis, which reverses the domain entirely. For f(x) = log(−x), you set −x > 0, which gives x < 0. The domain is (−∞, 0). The function only accepts negative x-values because the negative sign flips them positive before the log evaluates them.
When a reflection combines with a horizontal shift, you handle both at once. For f(x) = log₂(−x + 4), set −x + 4 > 0 and solve: −x > −4, so x < 4. The domain is (−∞, 4). The graph approaches a vertical asymptote at x = 4 from the left side.
Finding Domain With Quadratic or Rational Arguments
When the argument contains a quadratic expression, you need to solve a polynomial inequality instead of a simple linear one. For f(x) = ln(x² − 4), set x² − 4 > 0. Factor it as (x − 2)(x + 2) > 0, then use a sign chart or test values. The product is positive when both factors are positive (x > 2) or both are negative (x < −2). The domain is (−∞, −2) ∪ (2, ∞), two separate intervals.
If the argument is a fraction, you need the entire fraction to be positive, not just the numerator. For something like f(x) = log(x/(x − 1)), both the numerator and denominator affect the sign. You’d find where the fraction equals zero or is undefined, then test intervals to determine where the whole expression is positive.
Logarithms can also appear in denominators. For a function like g(x) = 1/ln(x), you need two things: x must be positive (so ln(x) is defined), and ln(x) cannot equal zero (to avoid dividing by zero). Since ln(1) = 0, you must exclude x = 1. The domain would be (0, 1) ∪ (1, ∞).
The Vertical Asymptote Shortcut
Every logarithmic function has a vertical asymptote where the argument equals zero. This boundary is also the edge of your domain. To find it, set the argument equal to zero and solve for x.
For f(x) = log₃(5x − 8), set 5x − 8 = 0 and solve to get x = 8/5. The vertical asymptote is the line x = 8/5, and the domain starts just to the right of it: (8/5, ∞). For f(x) = ln(7 − 2x), set 7 − 2x = 0 to get x = 7/2. Because of the negative coefficient on x, the domain extends to the left: (−∞, 7/2). The asymptote sits at the boundary in both cases.
Why the Range Is Almost Always All Real Numbers
The range of a logarithmic function is (−∞, ∞) for every standard case. This comes from the inverse relationship between logarithmic and exponential functions. The exponential function y = bˣ accepts all real numbers as inputs and produces only positive outputs. The logarithmic function, as its inverse, does the reverse: it accepts only positive inputs and produces all real numbers as outputs. Their domains and ranges swap.
This means that vertical transformations, which might seem like they’d limit the range, actually don’t. Multiplying the log by a constant, adding a number to it, or reflecting it upside down all stretch, shift, or flip the output values, but the outputs still cover every real number eventually. For example, f(x) = −log₂(x) still has range (−∞, ∞). So does f(x) = 5ln(x) + 12.
Horizontal transformations (changes inside the log) affect the domain but leave the range untouched. Vertical transformations (changes outside the log) would normally affect range, but since the range is already all real numbers, stretching or shifting infinity in any direction still gives you infinity. The range remains (−∞, ∞).
Step-by-Step Summary
For any logarithmic function, finding domain and range follows a consistent process:
- Identify the argument: the entire expression inside the log parentheses.
- Set the argument greater than zero: write the inequality and solve for x.
- Handle compound expressions carefully: for quadratics, use factoring and sign charts. For fractions, ensure the whole expression is positive and the denominator is nonzero.
- Write the domain in interval notation: always use parentheses on the boundary (never brackets), since the argument can approach zero but never reach it.
- State the range: for standard logarithmic functions and their transformations, the range is (−∞, ∞).
The base of the logarithm, whether it’s e, 10, 2, or any other positive number not equal to 1, never changes how you find the domain or range. Every logarithm follows the same restriction: the argument must be positive, and the output can be anything.