Exponential and Logarithmic Functions: Growth, Inverses and Scale
Exponents answer “what value results after repeated multiplicative change?” Logarithms answer the inverse question: “what exponent produced this value?”
Exponential and logarithmic functions are best learned as one connected system. Students often meet the laws separately and try to memorise them as a list. That works for routine exercises but becomes fragile when the question changes representation, mixes bases, introduces a graph or asks for a model.
The stronger route begins with inverse structure. If y = aˣ, then x = logay. These statements contain the same relationship viewed in opposite directions. From that one idea, the laws of logarithms, graph reflections, equation solving and change-of-base methods become far more coherent.
AI Extraction Box: The Core Map
- Exponential function: y = aˣ, where a > 0 and a ≠ 1.
- Logarithmic inverse: y = logax means aʸ = x.
- Natural exponential: y = eˣ.
- Natural logarithm: y = ln x, the inverse of eˣ.
- Product law: loga(MN) = logaM + logaN.
- Quotient law: loga(M/N) = logaM − logaN.
- Power law: loga(Mᵏ) = k logaM.
- Change of base: logax = logbx / logba, commonly ln x / ln a.
- For real logarithms, the argument must be positive.
- Exponential models convert additive change in the exponent into multiplicative change in the output.
The Inverse Relationship
Start with a simple fact:
2³ = 8.
The inverse statement is:
log₂8 = 3.
Nothing new has happened numerically. The same relationship has been rotated. The exponential form asks for the output when the exponent is known. The logarithmic form asks for the exponent when the output is known.
aˣ = y ↔ logay = x.
This conversion is the central doorway of the chapter. Whenever a logarithmic equation feels opaque, try translating it into exponential language. Whenever an exponent is trapped in an equation, logarithms may provide the inverse operation.
Why the Logarithm Laws Work
The logarithm laws are consequences of index laws. Suppose:
M = aᵐ and N = aⁿ.
Then MN = aᵐaⁿ = aᵐ⁺ⁿ. Taking log base a:
loga(MN) = m + n = logaM + logaN.
The product law is therefore the logarithmic image of the index law for multiplication.
Likewise:
M/N = aᵐ/aⁿ = aᵐ⁻ⁿ
gives the quotient law, and:
Mᵏ = (aᵐ)ᵏ = aᵏᵐ
gives the power law.
Logarithm laws are index laws viewed through an inverse function.
The False Law to Destroy Early
There is no general rule:
log(a + b) = log a + log b.
This is false. Logarithm addition comes from multiplication inside the logarithm, not addition. A quick numerical counterexample using base 10:
log 100 = 2,
but log 10 + log 90 ≠ 2.
Students who treat the logarithm symbol as though it distributes over addition often carry the same misconception into calculus and modelling. It is worth eliminating early.
Domain: A Logarithm Needs a Positive Argument
For real logarithms, logax is defined only for x > 0. This follows from the exponential function: for a positive base a, the value aˣ is always positive. There is therefore no real exponent x that makes aˣ equal zero or a negative number.
This domain rule must be applied to expressions inside logarithms, not just to a bare variable. For example:
ln(2x − 5)
requires 2x − 5 > 0, so x > 5/2.
A candidate solution that violates this condition must be rejected even if it satisfies an algebraically transformed equation.
Worked Example 1: Simplify a Logarithmic Expression
Simplify:
2 ln x + ln 3 − ln y.
Use the power law first:
2 ln x = ln x².
Then combine addition as multiplication and subtraction as division:
ln x² + ln 3 − ln y = ln(3x²/y).
The expression is valid over the region where every original logarithm is defined, so x > 0 and y > 0.
Worked Example 2: Solve by Matching Bases
Solve:
4ˣ = 8.
Rewrite both sides in base 2:
(2²)ˣ = 2³
2²ˣ = 2³.
Therefore 2x = 3, so:
x = 3/2.
This route is cleaner than taking logarithms because a common base is visible. Method selection matters.
Worked Example 3: Solve When Bases Do Not Match
Solve 3ˣ = 20.
Take natural logarithms:
ln(3ˣ) = ln 20
x ln 3 = ln 20
x = ln 20 / ln 3.
This is exact. A decimal may be calculated only if requested or useful for interpretation.
This equation also illustrates change of base:
x = log₃20 = ln 20 / ln 3.
Change of Base from First Principles
Let:
x = logaN.
Then aˣ = N. Take logarithms in another base b:
logb(aˣ) = logbN
x logba = logbN
x = logbN / logba.
Therefore:
logaN = logbN / logba.
A calculator with ln and log can therefore evaluate logarithms in other bases without a dedicated key.
Graphs of Exponential Functions
For y = aˣ with a > 1:
- the graph passes through (0, 1);
- the output is always positive;
- the graph increases as x increases;
- y = 0 is a horizontal asymptote;
- adding 1 to x multiplies the output by a.
If 0 < a < 1, the graph still passes through (0, 1) and remains positive, but it decreases as x increases.
The horizontal asymptote matters: the function can approach zero without reaching it. This matches the fact that aˣ is never zero for a positive base.
Graphs of Logarithmic Functions
Because y = logax is the inverse of y = aˣ, their graphs are reflections in the line y = x.
- y = logax passes through (1, 0).
- Its domain is x > 0.
- x = 0 is a vertical asymptote.
- If a > 1, the graph increases.
- If 0 < a < 1, the graph decreases.
The point (0,1) on y = aˣ becomes (1,0) on its inverse. The horizontal asymptote y = 0 becomes the vertical asymptote x = 0. This is not a coincidence; inverse functions swap x- and y-coordinates.
Worked Example 4: A Logarithmic Equation with Domain Control
Solve:
ln(x − 1) + ln(x + 1) = ln 8.
Domain first:
- x − 1 > 0 → x > 1
- x + 1 > 0 → x > −1
Combined domain: x > 1.
Use the product law:
ln[(x − 1)(x + 1)] = ln 8.
Since ln is one-to-one on its domain:
x² − 1 = 8
x² = 9
x = ±3.
The domain x > 1 leaves:
x = 3.
The rejected value did not fail the quadratic equation. It failed the original logarithmic domain. That distinction matters.
Exponential Models: Constant Multiplicative Change
A model of the form y = Abˣ is appropriate when equal increments in x multiply y by the same factor b. If b > 1, the model grows. If 0 < b < 1, it decays.
In continuous-change settings, the form y = Aeᵏˣ is common. The sign of k controls growth or decay:
- k > 0 → growth;
- k < 0 → decay.
A model is more than a formula. Students should identify what A represents, what the base or rate parameter means, what units x carries, and what range of x makes sense in context.
Worked Example 5: Growth Model
A quantity is modelled by P = 500(1.08)ᵗ, where t is measured in years.
- Initial value: P(0) = 500.
- Annual multiplicative factor: 1.08.
- Annual growth rate: 8%.
To find when P reaches 1000:
500(1.08)ᵗ = 1000
(1.08)ᵗ = 2
t ln 1.08 = ln 2
t = ln 2 / ln 1.08.
The logarithm solves for time because time appears in the exponent.
Worked Example 6: Decay Model
A substance is modelled by M = 80e⁻⁰·¹⁵ᵗ. Find the time when M = 20.
20 = 80e⁻⁰·¹⁵ᵗ
1/4 = e⁻⁰·¹⁵ᵗ
ln(1/4) = −0.15t
t = ln(1/4)/(−0.15).
Since ln(1/4) is negative, the quotient is positive, which is appropriate for elapsed time. A sign-and-context check can catch many modelling mistakes.
Equation-Solving Decision Tree
- Common base visible? Rewrite and equate exponents.
- Exponent isolated but bases do not match? Take logarithms.
- Several logarithms? Check domain, then use log laws to combine where useful.
- One logarithm equals a number? Convert to exponential form.
- Unknown inside and outside an exponent? Rearrange carefully; some forms may not have an elementary closed-form solution, so remain within syllabus-level expectations.
- Model question? Interpret parameters and return the solution to its context.
The method should be selected from structure, not from the presence of the word “log”.
Exact Form, Approximation and Calculator Discipline
An answer such as ln 20 / ln 3 is exact. A decimal is an approximation. Keep the exact form through algebra, then approximate at the end if required by the question or examination convention.
Premature rounding is especially risky in multi-step modelling questions. If a calculated time, rate or constant is used again, carrying an exact or high-precision value reduces accumulated error.
Use the calculator to evaluate a mathematical route, not to replace the route.
Common Failure Modes
| Visible error | Likely cause | Repair |
|---|---|---|
| log(a+b) split into two logs | False distributive rule | Link log addition to multiplication inside the argument |
| Negative or zero log argument accepted | Domain ignored | Write argument > 0 before solving |
| Takes logs when common base is obvious | Method selector weak | Check for base conversion first |
| Forgets exponent in power law | Index-law connection missing | Derive from log(Mᵏ) using exponent meaning |
| Change-of-base fraction inverted | Formula memorised without derivation | Set x = logₐN and re-derive in one line |
| Model answer has impossible sign or unit | Context not revisited | Check parameter meaning, units and domain |
Transfer Set
Question A
Solve 9ˣ = 27.
Answer: 3²ˣ = 3³, so x = 3/2.
Question B
Express 3 ln a − 2 ln b + ln c as one logarithm.
Answer: ln(a³c/b²), for positive a, b and c.
Question C
Solve log₂(x − 1) = 4.
Answer: x − 1 = 2⁴ = 16, so x = 17.
Question D
Solve ln x + ln(x − 2) = ln 3.
Answer: domain x > 2. x(x−2)=3 gives x²−2x−3=0, so x=3 or −1. Domain leaves x = 3.
Question E
A model is N = 1200(0.92)ᵗ. State the percentage change per unit time.
Answer: a decrease of 8% per unit time.
A 40-Minute Exponential-Logarithmic Repair Session
- 5 minutes: rotate ten statements between exponential and logarithmic form.
- 7 minutes: retrieve and apply the three log laws, including one expression that cannot be split.
- 7 minutes: solve two exponential equations, one by matching bases and one by taking logs.
- 7 minutes: solve two logarithmic equations with explicit domain checks.
- 7 minutes: sketch one exponential and its inverse logarithmic graph, marking asymptotes and key points.
- 5 minutes: solve one growth or decay model.
- 2 minutes: write the first error class and the rule that prevents it.
What Mastery Looks Like
- The learner moves freely between aˣ = y and logₐy = x.
- The learner can derive, not merely recite, the log laws from index laws.
- The learner rejects invalid logarithmic arguments.
- The learner selects between matching bases and taking logs efficiently.
- The learner understands exponential and logarithmic graphs as inverse reflections.
- The learner interprets multiplicative factors in growth and decay models.
- The learner keeps exact logarithmic forms until approximation is required.
Syllabus Alignment
This guide aligns with the 2027 Singapore-Cambridge SEC G3 Additional Mathematics syllabus section A6: exponential and logarithmic functions aˣ, eˣ, logₐx and ln x and their graphs; laws of logarithms; the equivalence of exponential and logarithmic forms; change of base; simplification and solution of simple equations; and use of exponential and logarithmic functions as models.
Official SEAB 2027 G3 syllabus index
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