DWG NO. 3.3 — Lesson 3 of 6

Logarithmic Functions as Inverses of Exponentials

Unit 3: Exponential and Logarithmic Functions · ~25 min

Objective Convert between exponential and logarithmic form, evaluate logarithms directly, and graph a logarithmic function as the inverse of an exponential one.

Back in 1.6, an inverse function undoes what the original function does, and its graph is a reflection of the original across y = x. Exponential functions are one-to-one, so every exponential function has an inverse — and that inverse is called a logarithm. The logarithm lo\(g_{b}\)(x) answers one question: "b to what power gives x?"

Definition

y = lo\(g_{b}\)(x) is equivalent to \(b^{y}\) = x

Same b, same relationship — just solved for the opposite variable. Every exponential equation can be rewritten as a logarithmic one, and vice versa.

Exponential form	Logarithmic form
2³ = 8	lo\(g_{2}\)(8) = 3
5² = 25	lo\(g_{5}\)(25) = 2
10⁻² = 0.01	lo\(g_{10}\)(0.01) = −2
b⁰ = 1	lo\(g_{b}\)(1) = 0

Two special bases

Common logarithm: lo\(g_{10}\)(x) is written simply as log(x) — base 10 is assumed when no base is shown.
Natural logarithm: lo\(g_{e}\)(x) is written ln(x) — the inverse of \(e^{x}\) from 3.2, and the most common logarithm in calculus.

Graphing a logarithm

Because lo\(g_{b}\)(x) is the inverse of \(b^{x}\), its graph is the exponential's graph reflected across the line y = x. Every feature swaps accordingly:

Feature	f(x) = \(b^{x}\)	f(x) = lo\(g_{b}\)(x)
Domain	(−∞, ∞)	(0, ∞)
Range	(0, ∞)	(−∞, ∞)
Intercept	(0, 1)	(1, 0)
Asymptote	horizontal, y = 0	vertical, x = 0

The vertical asymptote at x = 0 is the algebraic fingerprint of every logarithm: you can never take the log of zero or a negative number, because b raised to any real power is always positive.

y = 2ˣ and its inverse y = log₂x — reflections of each other across y = x

Worked Example 1 · Converting forms

ProblemRewrite 4³ = 64 in logarithmic form, and rewrite lo\(g_{3}\)(81) = 4 in exponential form.

1For 4³ = 64: base 4, exponent 3, result 64 → lo\(g_{4}\)(64) = 3.

2For lo\(g_{3}\)(81) = 4: base 3, the log equals 4, so 3 raised to 4 gives 81 → 3⁴ = 81.

lo\(g_{4}\)(64) = 3 · 3⁴ = 81

Worked Example 2 · Evaluating directly

ProblemEvaluate lo\(g_{5}\)(125) without a calculator.

1Ask: "5 to what power gives 125?" Since 5³ = 125, the exponent is 3.

lo\(g_{5}\)(125) = 3

Worked Example 3 · Domain of a log function

ProblemFind the domain of g(x) = lo\(g_{2}\)(x − 3).

1A logarithm is only defined when its argument is positive: x − 3 > 0.

2Solve: x > 3. The vertical asymptote of the parent lo\(g_{2}\)(x), normally at x = 0, has shifted right 3 units along with the graph.

Domain: (3, ∞) · vertical asymptote x = 3

Guided practice

1.Rewrite 6² = 36 in logarithmic form.
Show answer

lo\(g_{6}\)(36) = 2
2.Rewrite lo\(g_{2}\)(32) = 5 in exponential form.
Show answer

2⁵ = 32
3.Evaluate lo\(g_{3}\)(1).
Show answer

Any base raised to the power 0 gives 1, so 3⁰ = 1. lo\(g_{3}\)(1) = 0.
4.Evaluate lo\(g_{10}\)(0.001).
Show answer

10⁻³ = 0.001, so lo\(g_{10}\)(0.001) = −3.
5.State the domain and vertical asymptote of h(x) = ln(x + 5).
Show answer

Need x + 5 > 0, so x > −5. Domain: (−5, ∞) · vertical asymptote: x = −5.

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