QCE Mathematical Methods - Unit 3 - Differentiation of exponential and logarithmic functions

Calculus of logarithmic functions

Understand natural logarithms in QCE Mathematical Methods, including ln graphs, inverse relationships, log equations and logarithmic differentiation rules.

Part of the free QCE Maths Methods notes library for Unit 3: Differentiation of exponential and logarithmic functions.

Updated 2026-05-18 - 4 min read

The natural logarithm, $\ln(x)$, is the inverse of $e^x$. That sentence does a lot of work. It means $\ln(x)$ answers the question: "What power of $e$ gives me $x$?"

$ e^a=b \quad\Longleftrightarrow\quad \ln(b)=a $

Because $\ln(x)$ and $e^x$ are inverse functions, their graphs are reflections in the line $y=x$. It also explains why logarithms only accept positive inputs.

Log laws and conversion

Logarithms are another way of writing exponent equations:

$ a^x=y\quad\Longleftrightarrow\quad \log_a(y)=x. $

This is the fastest way to solve many equations. For example:

$ 2^{x-3}=9 $

becomes:

$ x-3=\log_2(9), $

so:

$ x=3+\log_2(9). $

The main laws are:

$ \log_a(xy)=\log_a(x)+\log_a(y), $

$ \log_a\left(\frac{x}{y}\right)=\log_a(x)-\log_a(y), $

and:

$ \log_a(x^k)=k\log_a(x). $

Also remember:

$ \log_a(1)=0,\quad \log_a(a^x)=x,\quad a^{\log_a x}=x. $

If no base is shown on a calculator-style $\log(x)$, the assumed base is usually $10$. Change of base lets you evaluate unfamiliar bases:

$ \log_a x=\frac{\log_b x}{\log_b a}. $

Graph features

For $y=\ln(x)$:

• domain: $x>0$
• range: all real numbers
• vertical asymptote: $x=0$
• $x$-intercept: $(1,0)$
• increasing, but gradually flattening

The key sketching idea is that the inside of the log must be positive. For $y=\ln(2x-4)$, the boundary is $2x-4=0$, so the vertical asymptote is $x=2$ and the domain is $x>2$.

Solving log equations

Use the inverse relationship to move between forms:

$ \ln(x)=3 \quad\Longleftrightarrow\quad x=e^3 $

If a question has multiple log terms, simplify with log laws first:

$ \ln(a)+\ln(b)=\ln(ab) $

$ \ln(a)-\ln(b)=\ln\left(\frac{a}{b}\right) $

$ k\ln(a)=\ln(a^k) $

These laws are powerful, but they can create extraneous solutions if you lose track of the domain.

Log equations often become quadratics after a substitution. For example:

$ 4^x-3(2^x)+2=0. $

Since $4^x=(2^x)^2$, let $u=2^x$:

$ u^2-3u+2=0. $

Then:

$ (u-1)(u-2)=0, $

so $2^x=1$ or $2^x=2$. Therefore $x=0$ or $x=1$. The substitution is not magic; it just makes the repeated exponential expression easier to see.

Differentiating ln

The core rule is:

$ \frac{d}{dx}\ln(x)=\frac{1}{x} $

For a function inside the logarithm:

$ \frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)} $

This is the chain rule. The derivative of the inside goes on top, and the original inside stays on the bottom.

For bases other than $e$:

$ \frac{d}{dx}\log_a x=\frac{1}{x\ln a}. $

This follows from change of base:

$ \log_a x=\frac{\ln x}{\ln a}. $

The denominator $\ln a$ is a constant. In Methods, most calculus questions use natural logarithms, but this formula explains why base $e$ is the cleanest base for differentiation.

Worked example

Domain checking example

Solve:

$ \ln(x)+\ln(x-2)=\ln(3) $

Combine the logs:

$ \ln(x(x-2))=\ln(3) $

So:

$ x(x-2)=3 $

$ x^2-2x-3=0 $

$ (x-3)(x+1)=0 $

The algebra gives $x=3$ or $x=-1$. But the original expression contains $\ln(x)$ and $\ln(x-2)$, so $x$ must be greater than $2$. The only valid solution is:

$ x=3 $

Logarithmic scales

Logarithmic scales are used when multiplication in the real quantity becomes addition on the scale. If:

$ L=\log_a(x)+b, $

then multiplying $x$ by a constant $c$ increases $L$ by:

$ \log_a(c). $

That is why decibels, pH, octaves and earthquake scales are logarithmic. For decibels:

$ L=10\log_{10}\left(\frac{I}{I_0}\right). $

If intensity is multiplied by $100$, the loudness level increases by:

$ 10\log_{10}(100)=20 $

decibels. The important modelling point is that equal jumps on the log scale represent equal multiplicative factors in the original quantity.

For pH:

$ \mathrm{pH}=-\log_{10}[\mathrm H^+]. $

A decrease of $1$ pH unit means the hydrogen ion concentration is multiplied by $10$, not merely increased by $1$.

Sketching transformed logarithms

For:

$ y=A\ln(kx+d)+c, $

the vertical asymptote comes from:

$ kx+d=0. $

The domain is the side of that asymptote where $kx+d>0$. To find an easy point, set the inside equal to $1$, because $\ln(1)=0$. For example, $y=\ln(2x-1)$ has asymptote $x=\frac12$ and crosses the $x$-axis when $2x-1=1$, so $x=1$.

Quick check

Sources

• QCAA Mathematical Methods subject page
• QCAA Mathematical Methods 2025 syllabus
• OpenStax Calculus Volume 1: Exponential and Logarithmic Functions
• OpenStax Calculus Volume 1: Derivatives of Exponential and Logarithmic Functions

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