Study Guide · Exponents & Logarithms

Exponents & Logarithms

Logarithms are just exponents written differently — once that clicks, all the laws follow.
6 sections 19 worked examples 13 checkpoints Intuition-first
1
Laws of Exponents
The rules that make everything else possible

An exponent tells you how many times a base is multiplied by itself. The laws below are not arbitrary rules — they follow directly from counting repeated multiplications. Understanding why each law works is more important than memorising it.

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Why do these laws work? x³ · x⁴ means (x·x·x)·(x·x·x·x) = x⁷ — you're just counting factors. Dividing cancels pairs. These laws are compressed notation for repeated multiplication.
LawFormulaExample
Productaᵐ · aⁿ = aᵐ⁺ⁿx³ · x⁴ = x⁷
Quotientaᵐ ÷ aⁿ = aᵐ⁻ⁿx⁵ ÷ x² = x³
Power of power(aᵐ)ⁿ = aᵐⁿ(x²)³ = x⁶
Zero exponenta⁰ = 17⁰ = 1
Negative exponenta⁻ⁿ = 1/aⁿx⁻³ = 1/x³
Fractional exponenta^(m/n) = ⁿ√(aᵐ)8^(2/3) = (∛8)² = 4
Roota^(1/n) = ⁿ√a27^(1/3) = 3
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Key rule: These laws only apply when the bases are identical. You cannot combine 2³ · 3² into a single power — the bases differ.
Medium
Combining exponent laws
Simplify: (2x²y)³ ÷ (4x⁵y²)
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Step 1 — Expand numerator: (2x²y)³ = 2³ · x⁶ · y³ = 8x⁶y³
Step 2 — Divide: 8x⁶y³ ÷ (4x⁵y²) = (8/4) · x^(6−5) · y^(3−2) = 2xy
2xy
Basic
Negative and zero exponents
Evaluate: (a) 5⁰   (b) 3⁻²   (c) (2/3)⁻³
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(a) Any non-zero base to the power 0 = 1 → 5⁰ = 1
(b) 3⁻² = 1/3² = 1/9
(c) (2/3)⁻³ = (3/2)³ = 27/8 = 3.375
5⁰ = 1, 3⁻² = 1/9, (2/3)⁻³ = 27/8
Medium
Fractional exponents
Simplify: (a) 27^(2/3)   (b) 16^(−3/4)   (c) (8x⁶)^(1/3)
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(a) 27^(2/3) = (³√27)² = 3² = 9
(b) 16^(−3/4) = 1/(16^(3/4)) = 1/(⁴√16)³ = 1/2³ = 1/8
(c) (8x⁶)^(1/3) = 8^(1/3) · (x⁶)^(1/3) = 2 · x² = 2x²
27^(2/3) = 9, 16^(−3/4) = 1/8, (8x⁶)^(1/3) = 2x²
Hard
Multi-step simplification
Simplify: (2a²b⁻³)³ ÷ (4a⁻¹b²)²
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Step 1 — Expand numerator: (2a²b⁻³)³ = 8a⁶b⁻⁹
Step 2 — Expand denominator: (4a⁻¹b²)² = 16a⁻²b⁴
Step 3 — Divide: (8a⁶b⁻⁹) ÷ (16a⁻²b⁴) = (1/2) · a^(6+2) · b^(−9−4) = (1/2)a⁸b⁻¹³
Step 4 — Positive exponents: a⁸ / (2b¹³)
a⁸ / (2b¹³)

Checkpoint 1 — Laws of Exponents

  1. Simplify: x⁴ · x⁻² · x³
  2. Evaluate: (−2)⁻⁴
  3. Simplify: (3m²n⁻¹)² ÷ (9m⁻²n)
  4. Write √(a⁵) using a fractional exponent
  1. x⁵ (4 − 2 + 3 = 5)
  2. 1/16 ((−2)⁴ = 16, then reciprocal)
  3. Numerator: 9m⁴n⁻²; divide by 9m⁻²n → m^(4+2) · n^(−2−1)/1 = m⁶/(3n³)
  4. a^(5/2)
2
Exponential Functions
y = a · cˣ — growth and decay

An exponential function has the variable in the exponent, not the base. This causes dramatically different behaviour compared to polynomials — exponentials grow (or decay) far faster.

Standard Form: f(x) = a · cˣ

  • a — initial value; the y-intercept is (0, a)
  • c — base; must be positive and not equal to 1
  • If c > 1: exponential growth (graph rises steeply)
  • If 0 < c < 1: exponential decay (graph falls toward 0)
  • Horizontal asymptote: y = 0 (the x-axis)
  • Domain: all real numbers; Range: y > 0 (when a > 0)
  • Alternative form: y = a(1 + r)ˣ where r is the rate (c = 1 + r)
Exponential growth (c > 1) vs exponential decay (0 < c < 1) y = 0 Growth: c = 1.12 Growth: c = 1.08 Decay: c = 0.88 (0, a)
Both growth curves start at (0, a). The decay curve falls toward y = 0 but never reaches it.
⚠️
Common mistake: The horizontal asymptote is always y = 0, not y = a. The y-intercept is a, but the asymptote is the x-axis.
Basic
Identify and evaluate an exponential function
For f(x) = 3 · 2ˣ: find f(0), f(3), f(−2); state the asymptote and range
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Evaluate:
f(0) = 3 · 2⁰ = 3 · 1 = 3
f(3) = 3 · 2³ = 3 · 8 = 24
f(−2) = 3 · 2⁻² = 3 · (1/4) = 3/4
Asymptote: y = 0    Range: y > 0
f(0) = 3, f(3) = 24, f(−2) = 3/4. Asymptote y = 0. Range: (0, +∞)
Medium
Find the rule from two points
f(x) = a · cˣ passes through (0, 5) and (2, 45). Find a and c.
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Step 1 — Use (0, 5): f(0) = a · c⁰ = a → a = 5
Step 2 — Use (2, 45): 5 · c² = 45 → c² = 9 → c = 3 (c must be positive)
f(x) = 5 · 3ˣ

Checkpoint 2 — Exponential Functions

  1. For f(x) = 4 · (1/3)ˣ, find f(0), f(1), f(−1)
  2. Is f(x) = 5 · 0.6ˣ growth or decay? What is the asymptote?
  3. Find the exponential function through (0, 2) and (3, 54)
  1. f(0) = 4, f(1) = 4/3, f(−1) = 12
  2. Decay (base 0.6 < 1). Asymptote: y = 0
  3. a = 2; 2·c³ = 54 → c³ = 27 → c = 3 → f(x) = 2·3ˣ
3
Growth & Decay Applications
Compound interest, half-life, and real-world models

Exponential models appear everywhere: population, money, radioactive decay, temperature cooling. The key is identifying the initial value and the rate, then applying the standard model.

General Model

  • Q(t) = Q₀ · (1 + r)ᵗ — growth if r > 0, decay if −1 < r < 0
  • Q₀ — initial quantity at t = 0
  • r — rate per period (e.g. 0.06 for 6% growth)
  • t — number of periods elapsed
  • Rule of 72: doubling time ≈ 72 ÷ (r%) years
Medium
Finding when an investment doubles
An investment of $1000 grows at 6% per year. After how many years does it double?
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Step 1 — Set up equation: 2000 = 1000(1.06)ᵗ → 2 = (1.06)ᵗ
Step 2 — Apply logarithm: log(2) = t · log(1.06)
Step 3 — Solve: t = log(2)/log(1.06) = 0.3010/0.0253 ≈ 11.9 years
Quick check: Rule of 72 → 72 ÷ 6 = 12 years ✓
t ≈ 11.9 years
Medium
Radioactive decay
A substance decays at 12% per year. Starting with 500 g, how much remains after 10 years?
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Step 1 — Identify parameters: Q₀ = 500, r = −0.12, c = 0.88, t = 10
Step 2 — Apply model: Q(10) = 500 · (0.88)¹⁰ ≈ 500 · 0.2785 ≈ 139 g
≈ 139 g remain after 10 years
Medium
Finding the rate from data
An investment grows from $1 200 to $1 680 in 4 years. Find the annual growth rate.
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Step 1 — Set up: 1200 · (1+r)⁴ = 1680 → (1+r)⁴ = 1.4
Step 2 — Take 4th root: 1+r = 1.4^(1/4) ≈ 1.0878
Step 3 — Solve: r ≈ 0.0878 → r ≈ 8.78%
Annual growth rate ≈ 8.78%
Hard
Carbon dating with half-life
Carbon-14 has a half-life of 5730 years. A fossil has 18% of original C-14. How old is it?
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Step 1 — Model: Q(t) = Q₀ · (0.5)^(t/5730)
Step 2 — Set up: 0.18 = (0.5)^(t/5730)
Step 3 — Take log: log(0.18) = (t/5730) · log(0.5)
Step 4 — Solve: t = 5730 · log(0.18)/log(0.5) ≈ 5730 · 2.474 ≈ 14 175 years
The fossil is approximately 14 175 years old

Checkpoint 3 — Growth & Decay

  1. A bacteria colony starts at 800 and triples every 6 hours. How many after 24 hours?
  2. A car depreciates 15% per year from $25 000. Value after 5 years?
  3. At what annual rate does money double in 8 years?
  1. 24 h = 4 periods → 800 · 3⁴ = 800 · 81 = 64 800
  2. 25 000 · (0.85)⁵ ≈ 25 000 · 0.4437 ≈ $11 093
  3. 2 = (1+r)⁸ → 1+r = 2^(1/8) ≈ 1.0905 → r ≈ 9.05%
4
What is a Logarithm?
The inverse of exponentiation

A logarithm answers the question: "To what power must I raise the base to get this number?" It is the inverse operation of exponentiation.

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The simplest way to think about logs: log₂(8) = ? asks "2 to WHAT POWER gives 8?" Answer: 3, because 2³ = 8. The logarithm is just the missing exponent.

Definition

  • logₐ(b) = c ⟺ aᶜ = b
  • Read: "log base a of b equals c"
  • Common log: log(x) means log₁₀(x)
  • Natural log: ln(x) means logₑ(x)
  • Change of base: logₐ(b) = log(b)/log(a)
  • Key values: logₐ(1) = 0 and logₐ(a) = 1
  • Domain: x > 0 always; Range: all real numbers
y = 2ˣ and its inverse y = log₂(x) — reflections across y = x y = x y = 2ˣ y = log₂(x) Key relationship: If y = 2ˣ, then x = log₂(y) Domain of 2ˣ: all reals Domain of log₂(x): x > 0
log₂(x) and 2ˣ are reflections of each other across y = x — they are inverse functions
💡
Bridge the definition: If you're stuck on any log problem, write it as bʸ = x. Convert, then solve the resulting exponential equation.
Basic
Evaluating logs from the definition
Find: log₃(81)
▶
Ask the question: "3 to what power gives 81?" → 3⁴ = 81
Answer: log₃(81) = 4
log₃(81) = 4
Basic
Convert between log and exponential form
(a) Write log₅(125) = 3 in exponential form   (b) Evaluate log₂(1/8)
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(a) log₅(125) = 3 ↔ 5³ = 125
(b) Ask: 2^? = 1/8 = 2⁻³ → log₂(1/8) = −3
(a) 5³ = 125    (b) −3
Medium
Evaluate logs with mismatched bases
Evaluate without a calculator: (a) log₄(8)   (b) log₉(27)
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(a) log₄(8): Write both as powers of 2: (2²)^y = 2³ → 2y = 3 → y = 3/2
(b) log₉(27): Write both as powers of 3: (3²)^y = 3³ → 2y = 3 → y = 3/2
Both equal 3/2

Checkpoint 4 — Logarithm Definition

  1. Convert to exponential form: log₇(49) = 2
  2. Evaluate log₃(1/27)
  3. For what x does log₅(x) = −1?
  4. Evaluate log₈(4)
  1. 7² = 49
  2. −3 (since 3⁻³ = 1/27)
  3. 5⁻¹ = 1/5, so x = 1/5
  4. 8^y = 4 → (2³)^y = 2² → 3y = 2 → y = 2/3
5
Laws of Logarithms
Three rules that mirror the exponent laws

The log laws exist because logs are exponents. The product rule mirrors aᵐ · aⁿ = aᵐ⁺ⁿ — logarithms of products become sums of logarithms. Mastering these lets you expand or condense any log expression.

The Three Log Laws (base b, with M, N > 0)

  • log_b(MN) = log_b(M) + log_b(N) — product rule
  • log_b(M/N) = log_b(M) − log_b(N) — quotient rule
  • log_b(Mⁿ) = n · log_b(M) — power rule
  • log_b(b) = 1 and log_b(1) = 0 — special values
  • log_b(M) = log(M)/log(b) — change of base formula
⚠️
The most common mistake: log(a + b) ≠ log(a) + log(b). There is no law for sums inside a log. Only products, quotients, and powers can be separated.
Exponent LawCorresponding Log Law
aᵐ · aⁿ = aᵐ⁺ⁿlog(MN) = log(M) + log(N)
aᵐ ÷ aⁿ = aᵐ⁻ⁿlog(M/N) = log(M) − log(N)
(aᵐ)ⁿ = aᵐⁿlog(Mⁿ) = n·log(M)
Basic
Expanding a logarithm
Expand: log(x²y/z³)
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Step 1 — Quotient rule: log(x²y) − log(z³)
Step 2 — Product rule on numerator: log(x²) + log(y) − log(z³)
Step 3 — Power rule: 2log(x) + log(y) − 3log(z)
2log(x) + log(y) − 3log(z)
Medium
Condensing to a single logarithm
Write as a single log: 3log(x) − 2log(y) + log(5)
▶
Step 1 — Power rule in reverse: log(x³) − log(y²) + log(5)
Step 2 — Product and quotient rules: log(5x³/y²)
log(5x³/y²)
Medium
Change of base
Calculate log₇(200)
▶
Formula: log₇(200) = log(200)/log(7)
Calculate: ≈ 2.3010/0.8451 ≈ 2.724
log₇(200) ≈ 2.724
Hard
Given log values, find combinations
Given log(2) ≈ 0.301 and log(3) ≈ 0.477, find: (a) log(6)   (b) log(1.5)   (c) log(72)
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(a) log(6) = log(2·3) = 0.301 + 0.477 = 0.778
(b) log(1.5) = log(3/2) = 0.477 − 0.301 = 0.176
(c) log(72) = log(8·9) = 3·log(2) + 2·log(3) = 0.903 + 0.954 = 1.857
(a) 0.778   (b) 0.176   (c) 1.857

Checkpoint 5 — Log Laws

  1. Expand: log₂(8x⁴/y)
  2. Condense: 3ln(x) − ln(x²) + ln(5)
  3. Compute log₃(50) using change of base
  4. Given log(5) ≈ 0.699, find log(0.04)
  1. log₂(8) + 4log₂(x) − log₂(y) = 3 + 4log₂(x) − log₂(y)
  2. ln(x³) − ln(x²) + ln(5) = ln(x³·5/x²) = ln(5x)
  3. log(50)/log(3) ≈ 1.699/0.477 ≈ 3.561
  4. log(0.04) = log(4/100) = 2log(2)−2. log(2)=1−0.699=0.301 → −1.398
6
Solving Equations
Exponential and logarithmic equations

Equations come in two types: the variable is in the exponent (exponential equation) or inside a log (logarithmic equation). Use the inverse operation to undo the function and isolate the variable.

Two Core Strategies

  • Method 1 — Same base: Write both sides with the same base, then equate exponents
  • Method 2 — Take log: If bˣ = k, then x = log(k)/log(b)
  • Method 3 — Convert log → exp: If log_b(x) = k, then x = b^k
  • Method 4 — Substitution: If it looks quadratic (e.g. 4ˣ − 5·2ˣ + 4 = 0), let u = 2ˣ
  • Always check solutions in log equations — argument must be > 0
⚠️
Extraneous solutions: When solving log equations, always verify your answer. If a solution makes any log argument ≤ 0, it must be rejected.
Basic
Same base method
Solve: 4^(x+1) = 8
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Step 1 — Write both sides base 2: 4^(x+1) = (2²)^(x+1) = 2^(2x+2);   8 = 2³
Step 2 — Equate exponents: 2x + 2 = 3 → x = 1/2
x = 1/2
Medium
Log both sides
Solve: 5ˣ = 30
▶
Step 1 — Take log of both sides: log(5ˣ) = log(30)
Step 2 — Power rule: x · log(5) = log(30)
Step 3 — Solve: x = log(30)/log(5) = 1.4771/0.6990 ≈ 2.113
x ≈ 2.113
Medium
Solve a logarithmic equation
Solve: log₃(2x − 1) = 4
▶
Step 1 — Convert to exponential: 3⁴ = 2x − 1 → 81 = 2x − 1
Step 2 — Solve: 2x = 82 → x = 41
Step 3 — Check: 2(41)−1 = 81 > 0 ✓ and log₃(81) = 4 ✓
x = 41
Medium
Combine logs before solving
Solve: log(x) + log(x − 3) = 1
▶
Step 1 — Product rule: log(x(x−3)) = 1
Step 2 — Convert: x(x−3) = 10 → x² − 3x − 10 = 0
Step 3 — Factor: (x−5)(x+2) = 0 → x = 5 or x = −2
Step 4 — Check: x = −2: log(−2) undefined → reject. x = 5: log(5)+log(2) = log(10) = 1 ✓
x = 5   (x = −2 is extraneous)
Hard
Exponential equation with different bases
Solve: 3^(2x) = 5^(x+1)
▶
Step 1 — Take log of both sides: 2x·log(3) = (x+1)·log(5)
Step 2 — Expand: 2x·log(3) = x·log(5) + log(5)
Step 3 — Collect x: x(2log(3) − log(5)) = log(5)
Step 4 — Solve: x = log(5)/(2log(3)−log(5)) ≈ 0.699/(0.954−0.699) ≈ 0.699/0.255 ≈ 2.74
x ≈ 2.74

Checkpoint 6 — Solving Equations

  1. Solve: 2^(3x−1) = 16
  2. Solve: 3ˣ = 50 (3 d.p.)
  3. Solve: log₅(3x + 2) = 3
  4. Solve: log(x + 4) − log(x) = 1 (check for extraneous solutions)
  5. Solve: 2^x · 4^(x+1) = 128
  1. 16 = 2⁴ → 3x−1 = 4 → x = 5/3
  2. x = log(50)/log(3) ≈ 1.699/0.477 ≈ 3.561
  3. 5³ = 3x+2 → 125 = 3x+2 → x = 123/3 = 41
  4. log((x+4)/x)=1 → (x+4)/x=10 → x+4=10x → x = 4/9. Check: both arguments positive ✓ → x = 4/9
  5. 2^x · (2²)^(x+1) = 2⁷ → 2^(3x+2) = 2⁷ → 3x+2=7 → x = 5/3
💡
Exam strategy: Before solving, ask: (1) Can I make the bases the same? (2) Is it a log equation to convert to exponential? (3) Do I need to combine logs first? Then solve — and always check log answers for validity.

Checkpoint 7 — Mixed Review

  1. Simplify: (x³y⁻²)⁴ / (x²y⁻¹)³ leaving no negative exponents
  2. A city has 120 000 people and grows 2.5% per year. After how many years does it reach 200 000?
  3. Condense: log(4) + 2log(3) − log(12)
  4. Solve: log₂(x+3) + log₂(x−3) = 4
  1. Numerator: x¹²y⁻⁸; Denominator: x⁶y⁻³; Result: x⁶y⁻⁵ = x⁶/y⁵
  2. 120000·(1.025)ᵗ = 200000 → t = log(5/3)/log(1.025) ≈ 0.2219/0.01072 ≈ 20.7 years
  3. log(4) + log(9) − log(12) = log(36/12) = log(3)
  4. log₂((x+3)(x−3)) = 4 → x²−9 = 16 → x² = 25 → x = 5 or −5. Check: x=−5 gives log₂(−2) → reject. Answer: x = 5