Topic 2.10: Inverses of Exponential Functions

📌 Big Idea

Logarithmic functions are the inverses of exponential functions. Everything that's true about inputs/outputs for exponential functions gets swapped for logarithmic functions.

General Form: f(x) = a · log_b x where b > 0, b ≠ 1, a ≠ 0

🔄 The Key Pattern Swap

⚡ Exponential Function

x values change additively (equally spaced)
y values change multiplicatively (constant ratio)

x	f(x)
1	2
3	4
5	8
7	16

x: +2 each step · y: ×2 each step

📊 Logarithmic Function

x values change multiplicatively (constant ratio)
y values change additively (equally spaced)

x	g(x)
2	1
4	3
8	5
16	7

x: ×2 each step · y: +2 each step

✏️ Example 1 — Identify: Exponential, Logarithmic, or Neither?

a) Exponential

x	h(x)
1	16
2	8
3	4
4	2

x: +1 (additive) · h(x): ÷2 (multiplicative) → Exponential

b) Logarithmic

x	k(x)
10	10
30	20
90	30
270	40

x: ×3 (multiplicative) · k(x): +10 (additive) → Logarithmic

c) Neither

x	p(x)
5	1
50	2
500	4
5000	8

x: ×10 · p(x): ×2 — both multiplicative → Neither

d) Logarithmic

x	q(x)
4	−1
8	−4
16	−7
32	−10

x: ×2 (multiplicative) · q(x): −3 (additive) → Logarithmic

Quick test: If x-values are equally spaced and y-values multiply → exponential. If x-values multiply and y-values are equally spaced → logarithmic. If both change multiplicatively → neither.

🔑 Exponential & Log as Inverse Pairs

If f(x) = bˣ and g(x) = log_b x, then f(g(x)) = g(f(x)) = x

They undo each other completely. The base must match.

✏️ Example 2 — Verify f(x) = 3ˣ and g(x) = log₃ x are Inverses

Show both compositions equal x

f(g(x)) = f(log₃ x) = 3^{log₃ x} = x ✓

g(f(x)) = g(3ˣ) = log₃(3ˣ) = x ✓

Both compositions equal x — confirming they are inverses. The key: 3 raised to log base 3 of x always collapses back to x.

✏️ Example 3 — Graphing the Inverse (k(x) = 2ˣ and k⁻¹(x) = log₂ x)

Reflection over y = x

To graph log₂ x: switch x and y coordinates of key points from 2ˣ, then connect.

k(x) = 2ˣ points

x	k(x)
−1	½
0	1
1	2
2	4
3	8

k⁻¹(x) = log₂ x points

x	k⁻¹(x)
½	−1
1	0
2	1
4	2
8	3

Graph facts: log₂ x passes through (1,0) and (2,1). It has a vertical asymptote at x = 0. Domain: x > 0. Range: all reals.

✏️ Example 4 — AROC of the Inverse

h(x) = aˣ contains (2,3) and (6,27). Find AROC of y = log_a x on [3,27].

Since y = log_a x is the inverse of h(x) = aˣ,
the points switch: h contains (2,3) and (6,27)
→ log_a x contains (3,2) and (27,6)

AROC = (log_a 27 − log_a 3) / (27 − 3)
       = (6 − 2) / (27 − 3)
       = 4 / 24
       = 1/6

Key step: The inverse swaps x and y. If (2,3) is on the exponential, then (3,2) is on the logarithm. Use these switched points for the AROC calculation.

Card 1 of 6

Concept

How do you tell a logarithmic function from an exponential function using a table?

Tap to reveal ✨

Answer

Exponential: x additive, y multiplicative. Log: x multiplicative, y additive.

The patterns are exactly swapped — because log and exponential are inverses of each other.

Topic 2.10

Inverses of Exponential Functions

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Question 1 of 5

A table shows x values 10, 30, 90, 270 and y values 10, 20, 30, 40. What type of function does this represent?

x: ×3 each step · y: +10 each step

Question 2 of 5

A table shows x values 5, 50, 500, 5000 and y values 1, 2, 4, 8. What type of function?

x: ×10 · y: ×2 — what do you notice?

Question 3 of 5

f(x) = 3ˣ and g(x) = log₃ x. What is f(g(x))?

Substitute g(x) = log₃ x into f: 3^(log₃ x) = ?

Question 4 of 5

k(x) = 2ˣ has the point (3, 8). What point must be on the graph of k⁻¹(x) = log₂ x?

Question 5 of 5

h(x) = aˣ contains (2, 3) and (6, 27). Find the AROC of y = log_a x over [3, 27].

The inverse swaps points: (2,3)→(3,2) and (6,27)→(27,6)

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Keep going!

Inverses ofExponential Functions

📌 Big Idea

Show both compositions equal x

Reflection over y = x

h(x) = aˣ contains (2,3) and (6,27). Find AROC of y = loga x on [3,27].

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Question Sent!

Inverses of
Exponential Functions

h(x) = aˣ contains (2,3) and (6,27). Find AROC of y = log_a x on [3,27].