Exponential and Logarithmic Functions

aˣ and its inverse logₐx: domains, ranges and graphs, the base-above-or-below-1 rule for inequalities, base conditions, and a self-inverse function.

Video coming soon

Exponential

2ˣ and (1/2)ˣ

aˣ needs a > 0, a ≠ 1: domain R, range (0, ∞), through (0, 1). Increasing for a > 1, decreasing for 0 < a < 1.

Logarithm

logₐx: domain (0, ∞), range R, through (1, 0). For a > 1 log inequalities keep their direction; for 0 < a < 1 they flip.

Self-inverse

Domain x < 1, range y < 1. Solving y = log(10 − 10ˣ) for x gives the same formula, so f⁻¹ = f and the graph is symmetric in y = x.

Worked examples

1. log₂(x − 1) < 3.

1 < x < 9.

2. Domain of log_(x − 1)(3 − x).

(1, 2) ∪ (2, 3).

3. f(x) = log₁₀(10 − 10ˣ). Find f⁻¹.

f⁻¹(x) = log₁₀(10 − 10ˣ), x < 1 (self-inverse).

JEE-style question

The solution set of log₀.₅(x − 2) > log₀.₅ 4 is:

(a)(2, 6)

(b)(6, ∞)

(c)(−∞, 6)

(d)(2, ∞)

Show the answer and the traps

Base 0.5 < 1 flips the sign: x − 2 < 4. The domain needs x − 2 > 0. So 2 < x < 6: option (a).

(b) forgets to flip. (c) forgets the domain, x > 2.

(d) keeps only the domain and ignores the inequality.

Watch out

Not flipping for a base between 0 and 1log base ½ is decreasing.

Ignoring base conditions for a variable baseThe base must be positive and not 1.

Forgetting the argument must be positivelog needs argument > 0.

Practice

1. Domain of log₂(x² − 4).

(−∞, −2) ∪ (2, ∞).

2. Solve log base 1/3 of x > 2.

0 < x < 1/9.

3. Range of 3^(x²).

[1, ∞).

4. Domain of log base x of (5 − x).

(0, 1) ∪ (1, 5).