Mathematics · Topic 1.2 · Extended
Logarithms & Exponential Functions
Understand exponential growth and decay, master the definition and laws of logarithms, and learn to solve exponential equations algebraically. This is Extended-level content essential for the eAssessment.
What You'll Learn
- Graph and interpret exponential functions f(x) = ax and their transformations
- Convert fluently between exponential and logarithmic form
- Apply all five laws of logarithms to simplify and evaluate expressions
- Solve exponential equations using logarithms
- Solve logarithmic equations and check solution validity
- Model real-world scenarios (compound interest, radioactive decay, population growth)
eAssessment Focus
Criterion A: Apply logarithm laws in multi-step, unfamiliar algebraic problems.
Criterion B: Prove logarithm laws from the definition and index laws.
Criterion C: Use correct notation (loga, ln, proper subscripts) and justify degree of accuracy.
Criterion D: Apply exponential/logarithmic models to real-world contexts (finance, science).
Mathematics · Topic 1.2
Exponential Functions
An exponential function has the variable in the exponent: f(x) = ax where a > 0 and a ≠ 1.
Key Properties of y = ax
| Property | When a > 1 (growth) | When 0 < a < 1 (decay) |
|---|---|---|
| y-intercept | (0, 1) | (0, 1) |
| Asymptote | y = 0 (x-axis) | y = 0 (x-axis) |
| Domain | All real numbers | All real numbers |
| Range | y > 0 | y > 0 |
| Behaviour as x → ∞ | y → ∞ (increases rapidly) | y → 0 (decreases toward asymptote) |
| Behaviour as x → −∞ | y → 0 | y → ∞ |
Transformations
- y = ax + k — vertical shift up by k; asymptote becomes y = k
- y = ax−h — horizontal shift right by h
- y = c · ax — vertical stretch by factor c; y-intercept becomes (0, c)
- y = a−x — reflection in the y-axis (growth ↔ decay)
The Number e
The constant e ≈ 2.71828... is the base of the natural exponential function y = ex. It arises naturally in continuous growth/decay models. The function ex has a special property: its rate of change (derivative) equals itself.
Key Fact: Every exponential function y = ax passes through (0, 1) because a° = 1 for all valid bases. The graph never touches the x-axis (the asymptote) — it only approaches it.
Mathematics · Topic 1.2
What is a Logarithm?
A logarithm answers the question: "To what power must I raise the base to get this number?"
Definition
loga(x) = y ⇔ ay = x (a > 0, a ≠ 1, x > 0)
Reading Logarithms
"log base a of x equals y" means "a raised to the power y gives x." The logarithm IS the exponent.
Examples — converting between forms:
| Exponential Form | Logarithmic Form |
|---|---|
| 2³ = 8 | log2(8) = 3 |
| 10² = 100 | log10(100) = 2 |
| 5° = 1 | log5(1) = 0 |
| 3−2 = 1/9 | log3(1/9) = −2 |
| e¹ = e | ln(e) = 1 |
Special Logarithms
Common Logarithm
log(x) = log10(x). Base 10 is the default on most calculators when you press "log".
Natural Logarithm
ln(x) = loge(x). Base e ≈ 2.718. Used extensively in science and calculus.
Key Values to Know
- loga(1) = 0 for any valid base (because a° = 1)
- loga(a) = 1 (because a¹ = a)
- loga(an) = n (logarithm and exponential are inverses)
- aloga(x) = x (inverse property)
Critical Rule: You CANNOT take the logarithm of zero or a negative number. The domain of loga(x) is x > 0. Always check that your solutions satisfy this constraint.
Mathematics · Topic 1.2
Laws of Logarithms
Each log law mirrors an index law. You must know them by name and be able to prove them from the definition.
Law 1 — Product Law
Formula
loga(mn) = loga(m) + loga(n)
Proof: Let loga(m) = p and loga(n) = q. Then ap = m and aq = n. So mn = ap × aq = ap+q (index law). Taking loga: loga(mn) = p + q = loga(m) + loga(n). ■
Law 2 — Quotient Law
Formula
loga(m/n) = loga(m) − loga(n)
Law 3 — Power Law
Formula
loga(mn) = n · loga(m)
This is the most important law for solving exponential equations — it "brings the exponent down."
Law 4 — Change of Base
Formula
loga(x) = logb(x)logb(a) = log(x)log(a) = ln(x)ln(a)
This lets you evaluate any logarithm using your calculator's log or ln button.
Law 5 — Inverse Properties
Formula
loga(ax) = x and aloga(x) = x
Summary Table
| Log Law | Formula | Related Index Law |
|---|---|---|
| Product | log(mn) = log m + log n | ap × aq = ap+q |
| Quotient | log(m/n) = log m − log n | ap ÷ aq = ap−q |
| Power | log(mn) = n log m | (ap)q = apq |
| Change of base | loga(x) = log(x)/log(a) | — |
| Inverse | loga(ax) = x | alogax = x |
Common Mistakes: log(m + n) ≠ log m + log n. There is NO law for the log of a sum. Also: log(m)/log(n) ≠ log(m/n). Don't confuse quotient of logs with log of a quotient.
Mathematics · Topic 1.2
Solving Exponential & Logarithmic Equations
Two main types: (1) exponential equations where the unknown is in the exponent, and (2) logarithmic equations where the unknown is inside a log.
Type 1: Solving Exponential Equations
- Isolate the exponential term on one side.
- Take logarithms of both sides (use log or ln).
- Apply the power law to bring the exponent down.
- Solve the resulting linear equation.
- Justify your degree of accuracy.
Worked Example — Solving Exponential Equations
3x = 20
given equation
log(3x) = log(20)
take log of both sides
x · log(3) = log(20)
power law
x = log(20)log(3) = 1.30100.4771
change of base
x ≈ 2.727 (3 s.f.)
Type 2: Solving Logarithmic Equations
- Combine logarithms using product/quotient/power laws.
- Convert to exponential form: loga(expression) = c ⇒ expression = ac.
- Solve the resulting equation.
- Check validity: ensure all arguments of log are > 0.
Worked Example — Solving Logarithmic Equations
log2(x) + log2(x − 3) = 2
given equation
log2[x(x − 3)] = 2
product law
x(x − 3) = 2² = 4
convert to exponential form
x² − 3x − 4 = 0
expand & rearrange
(x − 4)(x + 1) = 0
factorise
x = 4 or x = −1 → x = −1 invalid (log of negative)
check validity
x = 4
Real-World Applications
Compound Interest: A = P(1 + r/n)nt
To find t: take log of both sides and solve.
Half-life: N = N0 × (1/2)t/t½
To find t: log(N/N0) = (t/t½) × log(1/2).
To find t: take log of both sides and solve.
Half-life: N = N0 × (1/2)t/t½
To find t: log(N/N0) = (t/t½) × log(1/2).
eAssessment Tip: In Criterion D real-world problems, always justify your degree of accuracy. State WHY you round to a given number of significant figures (e.g., "I round to 3 s.f. because the input data is given to 3 s.f.").
Mathematics · Topic 1.2
Worked Examples
These examples show the multi-step reasoning expected at Grade 10 Extended level.
EXAMPLE 1Evaluate log2(32) without a calculator.
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Full Solution
I need to find the power to which 2 must be raised to give 32.
2¹ = 2, 2² = 4, 2³ = 8, 2&sup4; = 16, 2&sup5; = 32.
Therefore log2(32) = 5.
Check: 2&sup5; = 32 ✓
2¹ = 2, 2² = 4, 2³ = 8, 2&sup4; = 16, 2&sup5; = 32.
Therefore log2(32) = 5.
Check: 2&sup5; = 32 ✓
EXAMPLE 2Solve 52x−1 = 80. Give your answer to 3 significant figures.
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Full Solution
Take log of both sides:
(2x − 1) log 5 = log 80
2x − 1 = log 80 / log 5 = 1.9031 / 0.6990 = 2.7226...
2x = 3.7226...
x = 1.8613...
x ≈ 1.86 (3 s.f.)
Justification of accuracy: I round to 3 significant figures because the problem specifies this level of precision.
(2x − 1) log 5 = log 80
2x − 1 = log 80 / log 5 = 1.9031 / 0.6990 = 2.7226...
2x = 3.7226...
x = 1.8613...
x ≈ 1.86 (3 s.f.)
Justification of accuracy: I round to 3 significant figures because the problem specifies this level of precision.
EXAMPLE 3Simplify log3(27) + log3(9) − log3(81).
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Full Solution
Convert each to powers of 3:
log3(27) = log3(3³) = 3
log3(9) = log3(3²) = 2
log3(81) = log3(3&sup4;) = 4
Result: 3 + 2 − 4 = 1
Alternative using log laws: log3(27 × 9 / 81) = log3(243/81) = log3(3) = 1. ✓
log3(27) = log3(3³) = 3
log3(9) = log3(3²) = 2
log3(81) = log3(3&sup4;) = 4
Result: 3 + 2 − 4 = 1
Alternative using log laws: log3(27 × 9 / 81) = log3(243/81) = log3(3) = 1. ✓
EXAMPLE 4Prove that log(mn) = log m + log n using the definition of logarithms.
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Full Solution
Let log m = p and log n = q.
By definition: 10p = m and 10q = n.
Then mn = 10p × 10q = 10p+q (multiplication index law).
Taking log of both sides: log(mn) = log(10p+q) = p + q.
Since p = log m and q = log n:
log(mn) = log m + log n. ■
By definition: 10p = m and 10q = n.
Then mn = 10p × 10q = 10p+q (multiplication index law).
Taking log of both sides: log(mn) = log(10p+q) = p + q.
Since p = log m and q = log n:
log(mn) = log m + log n. ■
EXAMPLE 5Solve log5(2x + 3) = 2.
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Full Solution
Convert to exponential form:
2x + 3 = 5² = 25
2x = 22
x = 11
Check: log5(2(11) + 3) = log5(25) = log5(5²) = 2 ✓
Also: 2(11) + 3 = 25 > 0, so the argument is valid.
2x + 3 = 5² = 25
2x = 22
x = 11
Check: log5(2(11) + 3) = log5(25) = log5(5²) = 2 ✓
Also: 2(11) + 3 = 25 > 0, so the argument is valid.
EXAMPLE 6£5000 is invested at 3.5% annual interest, compounded annually. How many years until the investment exceeds £8000?
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Full Solution
A = P(1 + r)t where P = 5000, A = 8000, r = 0.035.
8000 = 5000(1.035)t
1.6 = (1.035)t
log(1.6) = t · log(1.035)
t = log(1.6) / log(1.035) = 0.2041 / 0.01494 = 13.66...
Since t must be a whole number of years and the investment must exceed £8000: t = 14 years.
Justification: I round UP to 14 because after 13 years the amount is still below £8000 (5000 × 1.03513 ≈ 7771), and after 14 years it exceeds £8000 (5000 × 1.03514 ≈ 8043).
8000 = 5000(1.035)t
1.6 = (1.035)t
log(1.6) = t · log(1.035)
t = log(1.6) / log(1.035) = 0.2041 / 0.01494 = 13.66...
Since t must be a whole number of years and the investment must exceed £8000: t = 14 years.
Justification: I round UP to 14 because after 13 years the amount is still below £8000 (5000 × 1.03513 ≈ 7771), and after 14 years it exceeds £8000 (5000 × 1.03514 ≈ 8043).
EXAMPLE 7Evaluate log4(8) using the change of base formula.
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Full Solution
Using change of base: log4(8) = log(8) / log(4)
= 0.9031 / 0.6021 = 1.5
Verification: 41.5 = 43/2 = (√4)³ = 2³ = 8. ✓
= 0.9031 / 0.6021 = 1.5
Verification: 41.5 = 43/2 = (√4)³ = 2³ = 8. ✓
Mathematics · Topic 1.2
Practice Q&A
Attempt each question before revealing the model answer. Show all working.
EVALUATEFind log3(243) without a calculator.
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Model Answer
243 = 3&sup5; (since 3 × 3 × 3 × 3 × 3 = 243).
log3(243) = log3(3&sup5;) = 5.
log3(243) = log3(3&sup5;) = 5.
SOLVESolve 4x = 50. Give your answer to 3 significant figures.
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Model Answer
x log 4 = log 50
x = log 50 / log 4 = 1.6990 / 0.6021 = 2.822...
x ≈ 2.82 (3 s.f.)
x = log 50 / log 4 = 1.6990 / 0.6021 = 2.822...
x ≈ 2.82 (3 s.f.)
SIMPLIFYWrite as a single logarithm: 2 log x + log y − 3 log z.
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Model Answer
Apply the power law: 2 log x = log(x²), 3 log z = log(z³).
Apply product law: log(x²) + log(y) = log(x²y).
Apply quotient law: log(x²y) − log(z³) = log(x²y / z³).
Apply product law: log(x²) + log(y) = log(x²y).
Apply quotient law: log(x²y) − log(z³) = log(x²y / z³).
SOLVESolve log(x) + log(x + 3) = 1. Check your solution(s).
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Model Answer
log[x(x + 3)] = 1 (product law, base 10)
x(x + 3) = 10¹ = 10
x² + 3x − 10 = 0
(x + 5)(x − 2) = 0
x = −5 or x = 2
Check: x = −5 → log(−5) is undefined. Invalid.
x = 2 → log(2) + log(5) = log(10) = 1 ✓
Solution: x = 2.
x(x + 3) = 10¹ = 10
x² + 3x − 10 = 0
(x + 5)(x − 2) = 0
x = −5 or x = 2
Check: x = −5 → log(−5) is undefined. Invalid.
x = 2 → log(2) + log(5) = log(10) = 1 ✓
Solution: x = 2.
APPLYA radioactive substance has a half-life of 12 hours. How long until only 10% of the original amount remains?
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Model Answer
N = N0 × (1/2)t/12. We need N/N0 = 0.10.
0.10 = (1/2)t/12
log(0.10) = (t/12) × log(0.5)
t/12 = log(0.10) / log(0.5) = (−1) / (−0.3010) = 3.322
t = 12 × 3.322 = 39.9 hours (≈ 40 hours to 2 s.f.)
0.10 = (1/2)t/12
log(0.10) = (t/12) × log(0.5)
t/12 = log(0.10) / log(0.5) = (−1) / (−0.3010) = 3.322
t = 12 × 3.322 = 39.9 hours (≈ 40 hours to 2 s.f.)
PROVEProve that loga(mn) = n · loga(m).
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Model Answer
Let loga(m) = p. Then ap = m (by definition).
mn = (ap)n = apn (power of a power index law).
Taking loga: loga(mn) = pn = n · p = n · loga(m). ■
mn = (ap)n = apn (power of a power index law).
Taking loga: loga(mn) = pn = n · p = n · loga(m). ■
EVALUATEFind log8(32) using the change of base formula.
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Model Answer
log8(32) = log(32) / log(8)
Both are powers of 2: 32 = 2&sup5;, 8 = 2³.
log(2&sup5;) / log(2³) = 5 log 2 / 3 log 2 = 5/3
Check: 85/3 = (2³)5/3 = 2&sup5; = 32 ✓
Both are powers of 2: 32 = 2&sup5;, 8 = 2³.
log(2&sup5;) / log(2³) = 5 log 2 / 3 log 2 = 5/3
Check: 85/3 = (2³)5/3 = 2&sup5; = 32 ✓
EXPLAINA student writes log(a + b) = log a + log b. Explain why this is incorrect.
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Model Answer
The product law states log(ab) = log a + log b, not log(a + b). There is no logarithm law for sums.
Counter-example: Let a = b = 5.
log(5 + 5) = log(10) = 1.
log(5) + log(5) = 2 × log(5) = 2 × 0.699 = 1.398.
1 ≠ 1.398, so the statement is false.
Counter-example: Let a = b = 5.
log(5 + 5) = log(10) = 1.
log(5) + log(5) = 2 × log(5) = 2 × 0.699 = 1.398.
1 ≠ 1.398, so the statement is false.
Mathematics · Review
Flashcard Review
Tap each card to reveal the answer. Try to answer from memory first.
What does loga(x) = y mean?
ay = x. "The power to which you raise a to get x."
Tap to reveal
State the product law of logarithms.
loga(mn) = loga(m) + loga(n)
Tap to reveal
State the quotient law of logarithms.
loga(m/n) = loga(m) − loga(n)
Tap to reveal
State the power law of logarithms.
loga(mn) = n · loga(m). It "brings the exponent down."
Tap to reveal
Change of base formula?
loga(x) = log(x) / log(a) = ln(x) / ln(a)
Tap to reveal
What is loga(1)?
0, for any valid base a. Because a° = 1.
Tap to reveal
What is loga(a)?
1. Because a¹ = a.
Tap to reveal
Can you take log of a negative number?
No. The domain of loga(x) is x > 0. Always check solutions are valid.
Tap to reveal
How do you solve ax = b?
Take log of both sides: x log a = log b, so x = log b / log a.
Tap to reveal
Does log(a + b) = log a + log b?
NO. There is no log law for sums. log a + log b = log(ab), not log(a + b).
Tap to reveal
What is e?
e ≈ 2.71828... The base of the natural logarithm (ln). It arises in continuous growth/decay.
Tap to reveal
What is an asymptote of y = ax?
The x-axis (y = 0). The graph approaches but never reaches it. y is always > 0.
Tap to reveal
What is the y-intercept of y = ax?
(0, 1) for all valid bases, because a° = 1.
Tap to reveal
How do you solve a log equation like log2(x) = 5?
Convert to exponential form: x = 2&sup5; = 32.
Tap to reveal
Compound interest formula?
A = P(1 + r/n)nt. To find t, take log of both sides and solve.
Tap to reveal
Mathematics · Practice Test
Practice Test — 20 Questions
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