Loav · VCE Mathematical Methods · Units 3 & 4
F04 · 2026 curriculum · Original teaching guide
1. The variable is in the exponent
An exponential function has a fixed positive base and a variable exponent. For y = bˣ, require b > 0 and b ≠ 1. The domain is ℝ and the range is (0, ∞). Every exponential value is positive, so an equation such as bˣ = −2 has no real solution. The graph passes through (0, 1), approaches the x-axis at one end, and never reaches it.
When b > 1, the function increases; when 0 < b < 1, it decreases. Equal additions to x multiply the output by a constant factor. This multiplicative structure distinguishes exponential change from linear change, where equal additions to x add a constant amount. To identify a model from equally spaced observations, compare ratios rather than only differences, while remembering that an added vertical offset changes those ratios.
The base e gives the natural exponential function eˣ. It obeys the same index laws as other positive bases. For example, eᵘeᵛ = e⁽ᵘ⁺ᵛ⁾ and eᵘ/eᵛ = e⁽ᵘ⁻ᵛ⁾. An exponent belongs to the whole base expression: e²ˣ means e raised to 2x, whereas 2eˣ means twice e raised to x. Use clear brackets in both written working and calculator entries.
2. A logarithm asks for an exponent
The statement logb(A) = c means bᶜ = A, with b > 0, b ≠ 1 and A > 0. A logarithm’s argument must be strictly positive, not merely non-negative. Logarithms are inverse functions of exponentials, so their domains and ranges swap: logb(x) has domain (0, ∞) and range ℝ. Its graph is the reflection of y = bˣ in y = x.
The natural logarithm ln(x) has base e. Thus ln(eˣ) = x for every real x, while eln(x) = x requires x > 0. The same inverse relationship lets us solve exponential equations when the unknown cannot conveniently be expressed as an integer exponent. A logarithmic answer is an exact number, not an unfinished calculation.
For positive A and B, the laws are ln(AB) = ln A + ln B, ln(A/B) = ln A − ln B, and ln(Aʳ) = r ln A when the stated real expressions are defined under this positive-base condition. There is no corresponding law that splits ln(A + B). The change-of-base relationship logb(A) = ln A/ln b follows by taking natural logarithms of bᶜ = A.
3. Isolate the exponential before taking logs
For an equation containing one exponential term, first move any added constants and divide by any outside coefficient. Check that the remaining right-hand side is positive. Then apply a logarithm to both sides and use the power law. Taking logarithms before isolating the exponential can create a logarithm of a sum, which cannot be separated into convenient pieces.
Worked example 1: an exact exponential solution
Solve 3e²ˣ − 7 = 5. Add 7 and divide by 3 to obtain e²ˣ = 4. The right-hand side is positive, so a real solution is possible. Taking natural logarithms gives 2x = ln 4, hence x = (ln 4)/2 = ln 2.
The equality of the two answer forms follows from ln 4 = ln(2²) = 2 ln 2. Substitution gives 3e2 ln 2 − 7 = 3 × 4 − 7 = 5. Because e²ˣ is strictly increasing, there is only one real solution. A decimal approximation is unnecessary unless the question asks for one.
When both e²ˣ and eˣ appear, a substitution can reveal a quadratic. Let u = eˣ, so u > 0 and e²ˣ = u². Solve the quadratic in u, discard any non-positive values, then solve eˣ = u for x. This is an example of reducing an unfamiliar-looking equation to a familiar structure while preserving the substitution’s range restriction.
4. Logarithmic equations begin with a domain line
List the positivity condition for every logarithm before combining terms. A combined logarithm can accidentally conceal invalid inputs. For instance, two negative factors have a positive product, but neither factor can serve as an individual real logarithm argument. The product law cannot legitimise those inputs. Solving a transformed equation produces candidates; the original domain decides which candidates survive.
Worked example 2: a plausible root that must be rejected
Solve ln(x − 1) + ln(x + 2) = ln 10. The separate conditions are x − 1 > 0 and x + 2 > 0, so the original domain is x > 1. Within that domain, combine the logarithms to obtain ln((x − 1)(x + 2)) = ln 10.
Natural logarithms are one-to-one, so (x − 1)(x + 2) = 10. Expanding gives x² + x − 12 = 0, or (x + 4)(x − 3) = 0. The candidates are −4 and 3. Reject −4 because it makes both original arguments negative. At x = 3, ln 2 + ln 5 = ln 10, so the only solution is x = 3.
5. Transformed graphs retain a boundary
For y = A eᵏˣ + C, with A and k non-zero, the domain is ℝ and the horizontal asymptote is y = C. Since eᵏˣ is positive, the range lies above C when A > 0 and below C when A < 0. The asymptote is not attained. The sign of k controls the exponential’s direction, while the sign of A can reflect that direction.
For y = A ln(B(x − H)) + K, with A and B non-zero, require B(x − H) > 0. If B is positive, the domain is x > H; if B is negative, it is x < H. The boundary x = H is a vertical asymptote. On the maximal domain, the range is ℝ. A restricted modelling domain can produce a smaller range, so separate the algebraic function from its contextual use.
6. Models: distinguish amount from excess above a limit
In N(t) = N₀eᵏᵗ, N₀ is the initial amount and k has units reciprocal to the time unit. Positive k represents growth and negative k decay. The doubling time for k > 0 is ln 2/k. The half-life for k < 0 is ln 2/(−k). These times are constant for the pure exponential model because the same proportional change repeats over equal time intervals.
With an offset, T(t) = C + Ae⁻ᵏᵗ for positive k, the exponentially decaying quantity is T − C, not T itself. The output approaches C without reaching it at a finite time in the model. Interpret units and the valid time interval, and avoid claiming the model will remain accurate indefinitely merely because its formula is defined there.
Worked example 3: time to approach a limiting temperature
A hypothetical model is T(t) = 18 + 62e−0.08t for t ≥ 0, with T in degrees Celsius and t in minutes. Initially T(0) = 80, and the model approaches 18 as time increases. Its contextual range is (18, 80], because time starts at zero.
For T = 30, solve 12 = 62e−0.08t. Hence t = ln(31/6)/0.08 ≈ 20.53 minutes. This time is non-negative and therefore admissible. The excess T − 18 halves after ln 2/0.08 ≈ 8.66 minutes; the temperature itself does not halve at each such interval. The values describe this proposed model, not measured experimental data.
Practise, then check
- Evaluate log2(1/8), eln 7, and ln(e⁻⁴) exactly.
- Solve log3(x − 2) = 2, stating the domain restriction first.
- Solve e²ˣ − 5eˣ + 6 = 0 exactly using a substitution.
- Solve ln(2x − 1) = ln(x + 4), checking the original arguments.
- For y = 2 − ln(x + 3), find the maximal domain, range, asymptote and both axis intercepts.
- A hypothetical population model is N(t) = 500e0.12t, t ≥ 0, with time in years. Find its initial population and doubling time.
- Solve ln(x − 2) ≤ ln 5. Explain why the solution does not include x = 2.
Full reasoned answers
- Since 2⁻³ = 1/8, the logarithm is −3. The inverse identities give eln 7 = 7 and ln(e⁻⁴) = −4. A negative logarithm value is allowed; a negative logarithm argument is not.
- Require x − 2 > 0, so x > 2. Converting to exponential form gives x − 2 = 3² = 9, hence x = 11, which meets the restriction.
- Let u = eˣ > 0. Then u² − 5u + 6 = (u − 2)(u − 3) = 0. Both u = 2 and u = 3 are positive, so x = ln 2 or ln 3. Each gives zero in the original equation.
- The domain requires x > 1/2 and x > −4, hence x > 1/2. Equality of natural logarithms gives 2x − 1 = x + 4, so x = 5. Both arguments then equal 9, confirming validity.
- Require x + 3 > 0, giving domain (−3, ∞). The range is ℝ and the vertical asymptote is x = −3. At y = 0, ln(x + 3) = 2, so the x-intercept is (e² − 3, 0). At x = 0, the y-intercept is (0, 2 − ln 3).
- N(0) = 500. For doubling, 1000 = 500e0.12t gives t = ln 2/0.12 ≈ 5.78 years. The rate parameter is 0.12 per year; this is not a statement that exactly 12% is added each full year.
- First require x > 2. Since ln is strictly increasing, ln(x − 2) ≤ ln 5 is equivalent to x − 2 ≤ 5 within that domain. Combining conditions gives 2 < x ≤ 7. At x = 2 the logarithm is undefined, so that endpoint stays open.
Exam habits and technology checks
Keep an exact logarithmic expression before giving an approximation, and round only the final requested quantity. Check that time units match the rate parameter. A numerical solution to a logarithmic equation must still satisfy every original argument condition. When using technology, record the mathematical equation and domain in ordinary notation rather than relying on calculator command syntax as your explanation.
Curriculum and sources
Scope: VCE Mathematical Methods Units 3 and 4, current 2023–2027 study design, Functions, relations and graphs and Algebra, number and structure. See the official VCAA Mathematical Methods curriculum and VCAA Mathematical Methods FAQs, updated February 2026. This guide covers exponential and logarithmic functions, inverses, equations and models.
Further reading: Michael Evans, Kay Lipson and David Greenwood, Cambridge Senior Mathematics VCE Mathematical Methods Units 3 & 4, second edition (2023), ISBN 9781009110495. Chapter 5, pp. 196–241. Explanations, diagrams, worked examples and practice questions in this guide are original. This is an independent Loav learning resource, not a VCAA publication.
