Additional Mathematics Synthesis Guide 15: Exponential and Logarithmic Change, Differentiation, Growth and Decay

Amount → relative rate → exponential model → logarithmic inverse → linearisation → interpretation.

1. Exponential functions multiply rather than add

For y = Ab^x with A > 0 and b > 0, each increase of 1 in x multiplies y by b:

y(x + 1)/y(x) = b.

This is fundamentally different from a linear function y = mx + c, where each increase of 1 in x adds the fixed amount m.

If b > 1, the model grows. If 0 < b < 1, it decays. If b = 1, the amount is constant.

2. The natural exponential is especially compatible with calculus

The function e^x has the property

d(e^x)/dx = e^x.

Therefore, for

y = Ae^kx,

the chain rule gives

y′ = kAe^kx = ky.

The derivative is proportional to the function itself. This is the defining differential relationship behind continuous exponential growth and decay.

3. Relative rate is constant

For y = Ae^kx,

y′/y = k.

This ratio is the instantaneous rate of change per unit amount. If x is time in years, k has units of per year. A positive k indicates growth; a negative k indicates decay.

Do not confuse a constant relative rate with a constant absolute rate. When k > 0, y′ = ky increases as y increases. The amount added per unit time becomes larger even though the proportional rate remains the same.

4. Initial conditions determine the scale factor

Suppose Q(t) = Ae^kt and Q(0) = 500. Then

500 = Ae⁰ = A.

So Q(t) = 500e^kt. The parameter A is the amount at t = 0 under this choice of time origin.

If instead the model is written with a shifted time variable, such as Q(t)=Be^k(t−3), then B is the amount at t = 3. A parameter’s interpretation depends on the representation used.

5. Worked model — determine the growth constant from data

A quantity is modelled by Q(t) = Ae^kt. Suppose Q(0) = 120 and Q(3) = 180. Find A and k.

From Q(0)=120, A=120. Then

180 = 120e^3k.

Thus e^3k = 3/2. Taking natural logarithms:

3k = ln(3/2), so k = (1/3)ln(3/2).

The exact model is

Q(t)=120e^[ln(3/2)/3]t.

6. The model predicts multiplicative consistency

For the model above, every 3 time units multiply the quantity by 3/2. Therefore

Q(6) = 120(3/2)² = 270.

This calculation can be done without first approximating k. The exponential model has preserved the interval growth factor exactly.

A real application should still ask whether the same growth mechanism is plausible over the extended interval. Mathematical consistency inside the model is not evidence that external conditions remain unchanged forever.

7. Doubling time depends only on k

For growth Q(t)=Ae^kt with k>0, the doubling time T satisfies

Ae^kT = 2A.

Cancel A and take logarithms:

T = ln2/k.

The initial amount does not appear. Under a constant relative-growth model, 100 takes the same time to become 200 as 500 takes to become 1000.

8. Half-life is the decay analogue

For decay Q(t)=Ae^kt with k<0, the half-life H satisfies

e^kH = 1/2.

Therefore

H = ln(1/2)/k = −ln2/k.

Since k is negative, H is positive.

9. Worked decay example

A quantity has half-life 8 hours and initial value 640. Under an ideal exponential-decay model, find the amount after 20 hours.

The decay constant is k = −ln2/8. Hence

Q(t)=640e^−(ln2)t/8 = 640(1/2)^t/8.

At t=20,

Q(20)=640(1/2)^5/2 = 80√2 ≈ 113.1.

Keeping the power form shows that 20 hours corresponds to two and a half half-lives.

10. Logarithms undo exponential relationships

The natural logarithm ln x is the inverse of e^x for x>0:

• ln(e^x)=x
• e^{ln x}=x for x>0

This is why logarithms are the natural tool when the unknown appears in an exponent.

Solve e^2x−1=7:

2x−1=ln7, so x=(1+ln7)/2.

11. Logarithm laws are multiplicative laws

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

These rules require positive logarithm arguments in real-number work. They do not imply

ln(a+b)=ln a+ln b.

Addition inside a logarithm usually cannot be separated in this way.

12. Domain control in logarithmic equations

Solve

ln(x−1)+ln(x+2)=ln4.

The original domain requires x>1. Combine the logarithms:

(x−1)(x+2)=4.

This gives x²+x−6=0, so x=2 or x=−3. Only x=2 belongs to the original domain.

The negative candidate satisfies the product equation but not the two original logarithms. The transformation has compressed two domain conditions into one algebraic product, so the original restrictions must remain visible.

13. Differentiating an exponential model

For

y=5e^2x,

the derivative is

y′=10e^2x=2y.

At x=0, the point is (0,5) and the tangent gradient is 10. Hence the tangent is

y−5=10x.

The normal gradient is −1/10, so the normal is y−5=−x/10.

14. Differentiating logarithms

For x>0,

d(ln x)/dx = 1/x.

Using the chain rule,

d[ln(3x+1)]/dx = 3/(3x+1),

where 3x+1>0 for the real logarithm.

The derivative formula does not remove the original domain. A derivative expression may itself be algebraically meaningful at points where the original logarithmic function was not defined; those points do not become part of the derivative’s function domain in the original problem.

15. A logarithmic model has diminishing absolute slope

Consider y = 4+3ln x for x>0. Then

y′=3/x.

The slope is positive but decreases as x increases. The function continues to rise, but equal increases in x produce smaller and smaller local changes in y.

This is qualitatively different from exponential growth, where y′ is proportional to y and therefore typically becomes larger as y grows.

16. Linearising an exponential model

Starting from

y=Ae^kx, A>0,

take natural logarithms:

ln y = kx + ln A.

Therefore a plot of ln y against x is linear under the model, with gradient k and vertical intercept ln A.

This connects exponential modelling to linear-law methods. A straight transformed graph supports the model over the observed range; it does not prove that the mechanism will remain exponential outside that range.

17. Worked linear-law example

Suppose a transformed graph of ln y against x is a straight line with gradient −0.4 and intercept ln50. Then

ln y = −0.4x + ln50.

Exponentiating gives

y = 50e^−0.4x.

The relative rate is −0.4 per unit x, and the model value at x=0 is 50.

18. Exponential equations can hide quadratics

Solve

e^2x−7e^x+12=0.

Let u=e^x, so u>0. Then

u²−7u+12=(u−3)(u−4)=0.

Thus e^x=3 or 4, giving

x=ln3 or ln4.

The exponential surface has revealed a quadratic structure. Guide 16 develops this hidden-structure technique further.

19. A calculus extension — optimise x e^−x

For x≥0, consider

f(x)=xe^−x.

Using the product rule,

f′(x)=e^−x(1−x).

Since e^−x>0, the sign of f′ is controlled by 1−x. The function increases for x<1 and decreases for x>1. Therefore its maximum on x≥0 occurs at x=1, with value

1/e.

This example shows why keeping the exponential factor visible is useful: it is always positive, so the derivative’s sign reduces to a simple linear factor.

20. Discrete percentage growth is not identical to continuous growth

A model that increases by 5% at the end of each discrete period has form

Q_n=Q₀(1.05)ⁿ.

A continuous model with parameter k has form Q(t)=Q₀e^kt. To give the same one-period multiplier 1.05, choose

e^k=1.05, so k=ln1.05.

The number 0.05 and the continuous parameter ln1.05 are close but not equal. The modelling convention matters.

21. Common failure patterns

• Adding a constant amount when the model is multiplicative.
• Calling k an absolute growth amount rather than a relative rate parameter.
• Using ln(a+b)=ln a+ln b.
• Ignoring the positive domain of logarithm arguments.
• Rounding k before exact cancellation is complete.
• Confusing doubling time with the time for an additive increase of A.
• Using a discrete percentage as though it were the continuous exponential parameter.
• Extrapolating an exponential model indefinitely without discussing its assumptions.

22. A reliable exponential-logarithmic routine

1. Identify whether change is additive or multiplicative.
2. State the model form and parameter meanings.
3. Use initial conditions to determine the scale factor.
4. Use logarithms when the unknown is in an exponent.
5. Carry logarithm domain restrictions.
6. Preserve exact logarithmic forms until approximation is required.
7. Use y′=ky to interpret continuous relative rate.
8. Return to the context and state where the model is being assumed valid.

23. Practice set

1. Differentiate 7e^3x.
2. Differentiate 5e^−2x.
3. Differentiate ln(4x+1).
4. Solve e^2x=5.
5. Solve 3e^x=11.
6. Solve ln(x−2)=ln5.
7. Solve ln(x−1)+ln(x+1)=ln8.
8. A model Q=200e^0.04t. What is Q(0)? What is Q′/Q?
9. Find its doubling time exactly.
10. A quantity follows Q=900e^−0.2t. Find its half-life exactly.
11. A quantity doubles every 6 time units. Find k in Q=Ae^kt.
12. A quantity has Q(0)=80 and Q(4)=120. Find k exactly.
13. Under Question 12’s model, find Q(8).
14. Rewrite ln y=−0.3x+ln40 as y=…
15. For y=6e^x, find the tangent at x=0.
16. For y=2+ln x, find the tangent gradient at x=e.
17. Solve e^2x−5e^x+4=0.
18. Find the maximum of xe^−x for x≥0.
19. What continuous k gives the same one-period multiplier as 8% discrete growth?
20. Why does a straight ln y against x graph support an exponential model?

Answers

1. 21e^3x.
2. −10e^−2x.
3. 4/(4x+1).
4. x=(1/2)ln5.
5. x=ln(11/3).
6. x=7.
7. Domain x>1; (x−1)(x+1)=8 gives x²=9, so x=3.
8. 200; 0.04 per unit t.
9. ln2/0.04=25ln2.
10. ln2/0.2=5ln2.
11. k=ln2/6.
12. e^4k=3/2, so k=(1/4)ln(3/2).
13. 80(3/2)²=180.
14. y=40e^−0.3x.
15. Point (0,6), slope 6, so y=6x+6.
16. 1/e.
17. Let u=e^x>0: (u−1)(u−4)=0, so x=0 or ln4.
18. At x=1, value 1/e.
19. k=ln1.08.
20. Because ln y=ln A+kx is linear in x under the model, with gradient k and intercept ln A.

24. What mastery looks like

Mastery means the learner can distinguish additive and multiplicative change, use logarithms as genuine inverses rather than formula decoration, interpret the derivative of an exponential model as a relative-rate statement, preserve domains, and move between nonlinear and linearised representations without losing parameter meaning.

The transfer test is to give a table rather than a formula, or a straight transformed graph rather than the original curve. If the learner can reconstruct the model, explain its parameters and state what assumptions would be required for extrapolation, the exponential system is becoming usable rather than memorised.

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• Guide 13: Radians, Arc Length, Sector Area and Trigonometric Modelling
• Guide 14: Rational Functions, Algebraic Fractions, Asymptotes and Domain Control
• Guide 16: Hidden Quadratics, Substitution, Structural Factorisation and Equation Families
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