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WAMath MethodsSyllabus dot point

How do we differentiate, integrate and apply exponential and logarithmic functions?

Differentiate and integrate exponential and natural logarithm functions and apply them to growth, decay and other modelling contexts

WACE Year 12 Mathematics Methods Unit 3 exponential and logarithmic functions: derivatives and integrals of e^x and ln x, the chain rule with exponentials, and growth and decay modelling with worked SCSA-style examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Derivatives
Integrals
Growth and decay modelling
Half-life and doubling time
Reading the rate from the model
Logarithm laws for solving

What this dot point is asking

SCSA Unit 3 develops calculus of the natural exponential function $e^{x}$ and the natural logarithm $\ln x$ . The defining feature of $e^{x}$ is that it is its own derivative, which makes it the natural model for any quantity whose rate of change is proportional to its current size. This dot point appears in both examination sections.

Derivatives

Worked derivative

For $y = e^{2x}\sin x$ apply the product rule with the chain rule on $e^{2x}$ :

\frac{dy}{dx} = 2e^{2x}\sin x + e^{2x}\cos x = e^{2x}(2\sin x + \cos x).

Integrals

Antidifferentiation reverses the derivatives above.

The absolute value in $\ln|x|$ matters because $1/x$ is defined for negative $x$ as well, but $\ln x$ is not.

Growth and decay modelling

A quantity whose rate of change is proportional to its size satisfies $\dfrac{dN}{dt} = kN$ , and the solution is the exponential model

N(t) = N_0 e^{kt},

where $N_0$ is the initial amount, $k>0$ gives growth and $k<0$ gives decay. Logarithms are used to solve for the time $t$ at which a target value is reached.

Radioactive decay

A sample decays with $N(t) = 80 e^{-0.1t}$ grams. When does it reach $20$ g?

Step 1: Set up the equation

We need to find the time $t$ at which $N(t)$ equals the target mass. Setting $N(t)=20$ and dividing both sides by $80$ isolates the exponential term.

80 e^{-0.1t} = 20 \;\Rightarrow\; e^{-0.1t} = \frac{20}{80} = 0.25.

Step 2: Apply the natural logarithm to solve for $t$

Taking $\ln$ of both sides removes the exponential because $\ln(e^{u})=u$ . The unknown then appears in a linear equation.

-0.1t = \ln(0.25) \;\Rightarrow\; t = \frac{\ln(0.25)}{-0.1} = \frac{-1.3863}{-0.1} \approx 13.9.

Final answer: The sample reaches $20$ g after approximately $13.9$ time units.

Half-life and doubling time

Two quantities are commonly asked of an exponential model. The doubling time of a growth model $N_0e^{kt}$ is found by setting $e^{kt}=2$ , giving $t=\dfrac{\ln 2}{k}$ . The half-life of a decay model is found by setting $e^{kt}=\tfrac{1}{2}$ , giving $t=\dfrac{\ln(1/2)}{k}=\dfrac{-\ln 2}{k}$ , which is positive because $k<0$ for decay. Both reduce to the same idea: take logarithms to pull the unknown out of the exponent. Because the result does not depend on $N_0$ , the doubling time and half-life are properties of the rate constant $k$ alone, so a population doubles in the same time no matter how large it already is.

Reading the rate from the model

The chain rule shows that for $N(t)=N_0e^{kt}$ the rate $N'(t)=kN(t)$ is always proportional to the current amount. This means a growth curve gets steeper as it rises (accelerating growth) while a decay curve flattens as it falls (slowing decline). Recognising the proportional-rate signature $\dfrac{dN}{dt}=kN$ tells you immediately that an exponential model applies and that $k$ is the constant of proportionality, a connection SCSA examines through both calculus and modelling questions.

Logarithm laws for solving

When solving exponential equations you will use:

\ln(ab) = \ln a + \ln b, \quad \ln\!\left(\frac{a}{b}\right) = \ln a - \ln b, \quad \ln(a^{k}) = k\ln a.

These let you bring an unknown out of an exponent before solving.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20217 marksCalculator-assumed. A population is modelled by

P(t)=2000e^{0.05t}

, where

t

is in years. (a) Find the initial population. (b) Find the rate of growth when

t=10

. (c) Find, to the nearest year, the time for the population to double.

Show worked answer →

An applied growth question combining differentiation and logarithms.

(a) At $t=0$ , $P(0)=2000e^{0}=2000$ .

(b) $P'(t)=2000\times 0.05\,e^{0.05t}=100e^{0.05t}$ . At $t=10$ , $P'(10)=100e^{0.5}\approx 165$ per year.

(c) Double means $P(t)=4000$ : $2000e^{0.05t}=4000$ , so $e^{0.05t}=2$ , giving $0.05t=\ln 2$ and $t=\dfrac{\ln 2}{0.05}\approx 13.9$ , about $14$ years.

Markers reward the initial value, the chain-rule derivative evaluated at $t=10$ , and using $\ln$ to solve for the doubling time.

WACE 20235 marksCalculator-free. (a) Differentiate

y=\ln(2x^{2}+1)

. (b) Evaluate

\displaystyle\int_{0}^{2} e^{-0.5x}\,dx

, giving an exact answer.

Show worked answer →

A calculator-free pairing of differentiation and integration.

(a) Chain rule $\dfrac{f'(x)}{f(x)}$ with $f(x)=2x^{2}+1$ : $\dfrac{dy}{dx}=\dfrac{4x}{2x^{2}+1}$ .

(b) $\displaystyle\int e^{-0.5x}\,dx=-2e^{-0.5x}+c$ . Apply the limits: $\bigl[-2e^{-0.5x}\bigr]_{0}^{2}=-2e^{-1}-(-2e^{0})=2-2e^{-1}=2-\dfrac{2}{e}$ .

Markers reward the chain rule in (a) and the $\dfrac{1}{k}$ factor with both limits substituted in (b).