Domain and Range of Logarithmic and Exponential Functions

The domain of a log function is not "all real numbers." That is the single most common mistake precalc students make, and it costs points on every exam. A logarithmic function f(x) = log_b(x) is defined only when its argument is strictly greater than zero. For the parent function f(x) = log_b(x), the domain is (0, ∞). Every shift and transformation you apply to the argument shifts that domain accordingly. If the argument is (x - h), the domain becomes (h, ∞). If the argument is a quadratic or rational expression, you must solve the inequality argument > 0 to find the domain.

An exponential function of the form f(x) = a·b^x + k has a domain of all real numbers (−∞, ∞). There is no value of x that makes the function undefined, so you never need to solve an inequality for the domain.

The range depends entirely on the vertical shift k and the sign of a. For the parent function f(x) = b^x with b > 0 and b ≠ 1, the range is (0, ∞). Adding k shifts that range to (k, ∞) if a > 0, or to (−∞, k) if a < 0. The line y = k is the horizontal asymptote. The function never reaches y = k; it gets arbitrarily close as x → −∞ (for a > 0) or as x → ∞ (for a < 0).

The failure case: students keep writing the range as (0, ∞) even after a vertical shift. If f(x) = 3·2^x + 5, the range is (5, ∞), not (0, ∞). The horizontal asymptote is y = 5, not y = 0. Check the sign of a: if a is negative, the range flips below the asymptote. For f(x) = −2·3^x + 4, the range is (−∞, 4).

For a logarithmic function f(x) = a·log_b(x - h) + k, the domain is found by setting the argument greater than zero: x - h > 0, so x > h. In interval notation, that is (h, ∞). The range of a logarithmic function is always all real numbers (−∞, ∞). No shift or stretch changes this. The range of exponential function is always restricted; the range of logarithmic function is always unrestricted.

Students frequently confuse the domain of the log with the domain of an exponential. The domain of log is restricted; the domain of the exponential is all reals. The range of the exponential is restricted; the range of the log is all reals. This swap is the direct result of the inverse relationship between the two families.

Quadratic Arguments

When the argument of a logarithm is not a simple linear expression, you must solve the inequality argument > 0. For f(x) = log(x² − 5x + 6), the domain is all x such that x² − 5x + 6 > 0. Factor the quadratic: (x − 2)(x − 3) > 0. Use a sign chart to determine where the product is positive. The zeros are at x = 2 and x = 3. Test intervals: for x < 2, both factors are negative, product positive. For 2 < x < 3, one factor negative and one positive, product negative. For x > 3, both factors positive, product positive. The domain is (−∞, 2) ∪ (3, ∞).

The failure case: students write the domain as "x > 2 or x > 3" or combine the two intervals incorrectly. The correct interval notation requires the union symbol. A common error is to write (2, 3) instead of the union of two separate intervals.

Rational Arguments

For a rational argument, such as f(x) = log((x+1)/(x−2)), solve (x+1)/(x−2) > 0. The fraction is zero at x = −1 and undefined at x = 2. Build a sign chart with these two critical points. The expression is positive when x < −1 and when x > 2. The domain is (−∞, −1) ∪ (2, ∞).

The exponential function f(x) = b^x and the logarithmic function g(x) = log_b(x) are inverses of each other. This means the domain of one is the range of the other. For f(x) = b^x, domain = (−∞, ∞) and range = (0, ∞). For g(x) = log_b(x), domain = (0, ∞) and range = (−∞, ∞).

When you shift either function, the swap applies to the shifted values. If f(x) = 2·3^x + 1 has range (1, ∞), then its inverse function g(x) = log_3((x−1)/2) has domain (1, ∞). The vertical shift of the exponential becomes the horizontal shift of the log. This relationship is the quickest way to check your work: if the domain of the log does not match the range of the exponential, you made an error.

Example 1: Exponential with Vertical Shift

Find the domain and range of f(x) = 4·5^x − 3. Domain: all real numbers (−∞, ∞). The horizontal asymptote is y = −3. Since a = 4 > 0, the range is (−3, ∞).

Example 2: Logarithm with Horizontal Shift

Find the domain and range of f(x) = 2·log_3(x + 5). Set the argument x + 5 > 0, so x > −5. Domain: (−5, ∞). Range: all real numbers (−∞, ∞).

Example 3: Log of a Quadratic Using a Sign Chart

Find the domain of f(x) = log(x² − x − 12). Factor: (x − 4)(x + 3) > 0. Zeros at x = −3 and x = 4. Sign chart: test x = −4 gives (−7)(−1) = 7 > 0; test x = 0 gives (−4)(3) = −12 < 0; test x = 5 gives (1)(8) = 8 > 0. Domain: (−∞, −3) ∪ (4, ∞).

Example 4: Exponential with Negative Coefficient

Find the domain and range of f(x) = −5·2^x + 10. Domain: all real numbers (−∞, ∞). The horizontal asymptote is y = 10. Since a = −5 < 0, the range is (−∞, 10). The function approaches 10 from below as x → −∞.

Function Family	Domain	Range	Asymptote
Exponential a·b^x + k, a > 0	All reals (−∞, ∞)	(k, ∞)	Horizontal: y = k
Exponential a·b^x + k, a < 0	All reals (−∞, ∞)	(−∞, k)	Horizontal: y = k
Log a·log_b(x − h) + k	(h, ∞)	All reals (−∞, ∞)	Vertical: x = h
Log of quadratic argument	Solve argument > 0 using sign chart	All reals (−∞, ∞)	Vertical asymptotes at zeros

This material suits precalculus students who have already mastered function notation from OpenStax College Algebra 2e, Section 3.1. If you can evaluate a function at a given x-value without error, you are ready for domain and range of log and exponential functions. Teachers and tutors preparing lessons will find the sign chart method for log of a quadratic and the horizontal asymptote exponential range rules directly usable.

Anyone looking for calculus-level continuity analysis or epsilon-delta proofs should go to a calculus textbook or Paul's Online Math Notes. Students who cannot evaluate a function at a given x-value should first master function notation before attempting domain and range work. The single thing that most often goes wrong: students forget that the argument of a logarithm must be strictly greater than zero, not greater than or equal to zero. A zero argument is undefined.