Additional Mathematics Synthesis Guide 2: Exponentials, Logarithms, Linear Law and Modelling

Multiplicative change → exponential model → logarithmic inverse → linearised graph → parameter estimate → return to the model.

1. Why exponential change is different from linear change

Linear change adds the same amount during equal steps. Exponential change multiplies by the same factor during equal steps. If a quantity increases by 5 units each hour, the difference is constant. If it increases by 5% each hour, the ratio between successive values is constant.

A discrete exponential model can be written as y = Ab^x. A continuous model is often written as y = Ae^kx. In both forms, A controls the scale or initial value, while b or k controls the multiplicative rate.

• b > 1 gives growth in y = Ab^x.
• 0 < b < 1 gives decay.
• k > 0 gives growth in y = Ae^kx.
• k < 0 gives decay.

2. The laws of indices carry the structure

Exponentials are manageable because multiplication of like bases can be compressed into addition of powers:

• a^maⁿ = a^m+n
• a^m/aⁿ = a^m−n
• (a^m)ⁿ = a^mn
• a⁰ = 1 for a ≠ 0
• a⁻ⁿ = 1/aⁿ

These are not arbitrary rules. They preserve the meaning of repeated multiplication. When the algebra later becomes dense, returning to that meaning is often safer than trying to recall a rule in isolation.

3. Logarithms answer the inverse question

If a^x = N, then log_aN = x. A logarithm asks: what power of a produces N?

For example, 2⁵ = 32 means log₂32 = 5. The two statements contain the same relationship in opposite directions.

The logarithm laws mirror the index laws:

• log(ab) = log a + log b
• log(a/b) = log a − log b
• log(aⁿ) = n log a

The important restriction is that real logarithms require a positive argument. A candidate solution that makes a logarithm of zero or a negative number is invalid even if it survives the algebra.

4. Worked Example 1 — solve by matching bases

Solve 3^2x−1 = 27.

Since 27 = 3³, equate powers:

2x − 1 = 3, hence x = 2.

This route is efficient because both sides can be expressed with the same base. Logarithms are available, but they are not necessary.

5. Worked Example 2 — solve when matching bases is inconvenient

Solve 5^x = 17.

Take natural logarithms:

x ln 5 = ln 17

so x = ln 17 / ln 5.

The exact form should normally be preserved until a decimal is actually required.

6. Worked Example 3 — logarithmic equation with a domain check

Solve log₂(x − 1) + log₂(x + 3) = 3.

First note the domain: x − 1 > 0 and x + 3 > 0, so x > 1.

Combine the logarithms:

log₂[(x − 1)(x + 3)] = 3.

Therefore

(x − 1)(x + 3) = 8

x² + 2x − 11 = 0.

Hence x = −1 ± 2√3. Only x = −1 + 2√3 satisfies x > 1.

7. Linear law: making a curved relationship straight

A straight-line equation has the form Y = mX + c. Linear law problems ask us to choose transformed variables X and Y so that a non-linear original relationship becomes straight.

For y = Ab^x, take natural logarithms:

ln y = ln A + x ln b.

If we plot Y = ln y against X = x, the gradient is ln b and the vertical intercept is ln A.

The graph is straight not because the original phenomenon became linear, but because we changed coordinates to expose its multiplicative structure.

8. Power laws can be linearised too

Suppose y = axⁿ. Taking logarithms gives

ln y = ln a + n ln x.

Plot Y = ln y against X = ln x. The gradient is n and the intercept is ln a. This is a powerful general pattern: a power becomes a multiplier after taking logs.

9. Worked Example 4 — recover an exponential model from a straight line

A graph of ln y against x is a straight line with gradient 0.4 and vertical intercept 1.2. Find the model in the form y = Ae^kx.

Since

ln y = kx + ln A,

we identify k = 0.4 and ln A = 1.2. Therefore A = e^1.2.

The model is y = e^1.2e^0.4x = e^{1.2 + 0.4x}.

10. Worked Example 5 — recover a power law

A graph of ln y against ln x has gradient 3 and vertical intercept −2. Find y in terms of x.

Compare

ln y = n ln x + ln a

with Y = 3X − 2. Thus n = 3 and ln a = −2, so a = e⁻².

Therefore y = e⁻²x³.

11. Growth and decay models need interpretation

Suppose N(t) = 500e^−0.12t. The initial value is N(0) = 500. The negative exponent means the model decays. To find when the quantity falls to 200:

200 = 500e^−0.12t

0.4 = e^−0.12t

ln 0.4 = −0.12t

t = −ln 0.4 / 0.12.

The logarithm enters because time is in the exponent and must be brought down into an ordinary multiplicative position.

12. Half-life and doubling time emerge from the same structure

For decay N = N₀e^−kt, half-life T satisfies

1/2 = e^−kT.

Taking logs gives T = ln 2 / k.

For growth N = N₀e^kt, doubling time is also ln 2 / k. The formula is not something separate to memorise; it follows directly from the model.

13. Why transformed graphs can mislead if interpretation is careless

A good straight line on transformed axes is evidence that a model is plausible over the observed range. It is not proof that the relationship must continue forever. Exponential growth in a real system may eventually encounter limited resources. Exponential decay may approach a background level not represented by the simplest model. A power law may describe one regime but not another.

Mathematics becomes modelling only when the student returns from the transformed graph to the original quantities and asks whether the conclusion remains meaningful.

14. Calculator control matters here

• Keep enough significant figures in intermediate values.
• Use brackets around complete logarithm arguments.
• Distinguish ln from log base 10 when the algebra assumes a particular base.
• Preserve exact expressions such as ln 7 / ln 5 until the last stage when possible.
• Check that the final value satisfies the original equation and any logarithm-domain restriction.

15. Common failure patterns

• Treating percentage growth as addition. Repeated percentage change is multiplicative.
• Applying log(a + b) = log a + log b. This is false; logarithm laws apply to products, quotients and powers.
• Forgetting domain restrictions. Logarithm arguments must be positive.
• Linearising but forgetting what X and Y mean. The gradient and intercept must be translated back into original parameters.
• Reading the intercept as A when it is actually ln A.
• Premature rounding. Small errors in a logarithmic parameter can propagate through an exponential model.
• Accepting an algebraic answer without checking the original model.

16. A reliable decision routine

1. Ask whether change is additive or multiplicative.
2. Identify the exponential or power-law form.
3. If the unknown is in an exponent, consider logarithms.
4. Before solving a log equation, state the domain.
5. For linear law, write the target straight-line form Y = mX + c explicitly.
6. Choose transformed variables and identify gradient and intercept.
7. Recover the original parameters.
8. Substitute back or otherwise verify.
9. Interpret the result in the original context.

17. Practice set

1. Solve 5^x+1 = 125.
2. Solve log₃(x − 1) = 2.
3. Solve ln x + ln(x − 3) = ln 10.
4. Express log₂10 using natural logarithms.
5. A graph of ln y against x has gradient 0.2 and intercept ln 4. Find y in terms of x.
6. State suitable transformed variables to linearise y = axⁿ.
7. A graph with Y = ln y and X = ln x has equation Y = 3X − 2. Find the original relation.
8. A quantity starts at 200 and decreases by 8% each equal time step. Write a discrete model after n steps.
9. Solve e^2x = 7 exactly.
10. For N = 80e^−0.3t, find the half-life.
11. If y = 5·2^x, by what factor does y change when x increases by 3?
12. Why must a logarithmic candidate solution be checked in the original domain?

Answers

1. x = 2.
2. x = 10.
3. x = 5.
4. ln 10 / ln 2.
5. y = 4e^0.2x.
6. X = ln x and Y = ln y, giving Y = nX + ln a.
7. y = e⁻²x³.
8. 200(0.92)ⁿ.
9. x = (1/2)ln 7.
10. T = ln 2 / 0.3 ≈ 2.31 time units.
11. A factor of 8.
12. Because algebra can produce values for which one or more logarithm arguments are zero or negative, where the original expression is undefined.

18. What mastery looks like

A student has moved beyond isolated procedures when they can distinguish additive from multiplicative change, move between exponential and logarithmic forms, control domains, recognise when a logarithm is the efficient inverse operation, linearise a model deliberately, read the transformed gradient and intercept correctly, recover the original parameters and explain what those parameters mean.

The strongest test is transfer. Give the learner a relationship whose variables and letters have changed. Ask which transformation would make it linear. Ask what the graph’s slope means in the original model. If those decisions remain available without a memorised template, the structure is becoming durable.

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Continue through the BTT Additional Mathematics system

• Additional Mathematics Directory.
• BTT Singapore Mathematics Hub.
• Guide 1: Quadratics, Coordinate Geometry and Tangency.
• Guide 3: Trigonometric Functions, Identities, Equations and Geometry.
• Guide 4: Calculus, Algebra and Kinematics.