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

Calculus of Logarithmic Functions | QCE Mathematical Methods

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

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