Exponential and logarithmic functions: WACE Year 12 Mathematics Methods Units 3 and 4

Syllabus placement. In the SCSA Mathematics Methods ATAR Year 12 syllabus, the logarithmic function (including its derivative) is Unit 4 (Topic 4.1), not Unit 3. Derivatives of exponential functions are Unit 3. This page sits in our Unit 3 folder for historical reasons.

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 $e^{2x}$:

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

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.

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:

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