Exponential and Logarithmic Functions

Exponential functions model repeated multiplication; logarithms undo it. This lesson builds on exponent rules in Algebra and inverse functions in Functions.

Growth, Decay, and the Number e

In f(x) = ab^x, a is the initial value. If b > 1 there is growth; if 0 < b < 1 there is decay. The domain is all real numbers, the range is positive when a > 0, and y = 0 is a horizontal asymptote.

Percent growth: A = P(1 + r)ᵗ
Percent decay:  A = P(1 − r)ᵗ
Continuous change: A = Peᵏᵗ

Worked example: compound growth

A $2,400 investment grows by 4.5% per year. Find its value after 6 years.

A = 2400(1 + 0.045)⁶
  = 2400(1.045)⁶
  ≈ 2400(1.302260...)
  ≈ $3,125.42

The answer is larger than $2,400, as a positive growth rate requires.

Logarithms Undo Exponents

The statements b^y = x and log_b(x) = y are equivalent. A logarithm answers, “What exponent on the base gives this number?” The base must be positive and not 1; the logarithm input must be positive.

Exponential form	Logarithmic form
25 = 32	log2(32) = 5
10−3 = 0.001	log(0.001) = −3
e2 = x	ln x = 2

Logarithm Laws

log_b(MN) = log_b M + log_b N
log_b(M/N) = log_b M − log_b N
log_b(Mᵖ) = p log_b M
change of base: log_b M = log M / log b

There is no sum law: log(M + N) cannot be split.

Worked example: solve an exponential equation

Solve 3^2x−1 = 20.

ln(3²ˣ⁻¹) = ln 20
(2x − 1)ln 3 = ln 20
2x − 1 = ln 20 / ln 3
2x = 1 + ln 20 / ln 3
x = (1 + ln 20 / ln 3)/2
x ≈ 1.8636

Check by substitution: 3^{2(1.8636)−1} ≈ 20.

Worked example: solve a logarithmic equation

Solve log₂(x − 1) + log₂(x − 3) = 3.

Domain first: x − 1 > 0 and x − 3 > 0, so x > 3.

log₂[(x − 1)(x − 3)] = 3
(x − 1)(x − 3) = 2³
x² − 4x + 3 = 8
x² − 4x − 5 = 0
(x − 5)(x + 1) = 0
x = 5 or x = −1

Only x = 5 satisfies x > 3.

Applications: Half-Life

A 160 mg sample has a half-life of 12 hours. When will 20 mg remain?

20 = 160(1/2)ᵗ⁄¹²
1/8 = (1/2)ᵗ⁄¹²
(1/2)³ = (1/2)ᵗ⁄¹²
3 = t/12
t = 36 hours

Practice Set

1. A population of 800 grows by 3% annually. Write a model and find the population after 5 years.
2. Rewrite 5³ = 125 in logarithmic form.
3. Expand ln[(x³√y)/z], assuming all variables are positive.
4. Solve 7^x = 50 to three decimal places.
5. Solve ln(x − 2) = 1 and check the domain.

Answer Checks

1. P(t) = 800(1.03)^t; P(5) ≈ 927.4, so about 927 individuals.
2. log₅(125) = 3.
3. 3 ln x + (1/2) ln y − ln z.
4. x = ln 50/ln 7 ≈ 2.010.
5. x − 2 = e, so x = e + 2. This is greater than 2, so it is valid.