Unit 3 · Topic 1.2 · Differentiation of exponential and logarithmic functions
Logs are the other half of Topic 1. In this guide you will graph them, solve equations with them, and differentiate them, with the real QCAA questions for each skill. Work through it in order, and tick off each skill once you can do it without help.
Calculus of logarithmic functions, about 8 hours of class time. Only tick a skill when you could do it in an exam.
In the old syllabus, logs in bases other than $e$ were part of Unit 3. In the new syllabus they have moved to Unit 2. The QCAA can still expect you to use Unit 1 and 2 knowledge in the external, but they aim to examine Unit 3 and 4 content directly.
So I would expect base $e$ to be the main log you meet in your external. Other bases used to be a big part of early exams, which is why some of the past questions below still use them.
Part 1
$y=\ln x$ is the inverse of $y=e^x$. Every point $(a,b)$ on one graph becomes $(b,a)$ on the other, so the graph of $\ln x$ is $e^x$ reflected in the line $y=x$. Its horizontal asymptote turns into a vertical one, and its $y$-intercept $(0,1)$ turns into the $x$-intercept $(1,0)$.
Watch it first
Because they undo each other, $e^{\ln x}=x$ and $\ln(e^x)=x$. You will use that in Part 2 to solve equations.
Just like exponentials, plotting a point on a log graph beyond the $x$-intercept needs a calculator or an estimate. That leaves the exam writers stuck: it is too easy in a technology-active paper and hard to mark in a technology-free one. So I think it is unlikely you will be asked to graph one directly.
What matters more is knowing the general form, $y=a\ln(x-h)+k$, and the four basic shapes well enough to pick the right graph in a multiple choice question. In short: a negative $a$ reflects the graph in the $x$-axis, and a negative in front of $x$ reflects it in the $y$-axis.
The dashed line is the vertical asymptote, $x=0$, in every case.
Watch it first
Most graph-to-equation questions come down to two features: the vertical asymptote and the $x$-intercept. Watch how each one gives you an unknown, then try it yourself below.
From a graph to an equation
Try it yourself
Move the sliders and watch the two features. The asymptote only depends on $h$. The $x$-intercept is where $a\ln(x-h)+k=0$, so it moves with all three.
Practise
Some of these are not in base $e$. The three steps above still work whatever the base, so I have included them to show you how the QCAA approach these questions.
For the written-answer questions, the solutions show where each mark goes, based on the QCAA marking guides. Many marks allow follow-through: if an earlier step is wrong but you use your wrong answer correctly, you can still get the later marks. That is another reason to show every step.
Which graph represents the function $f(x)=-3-\ln(x+3)$?
Asymptote first: $x+3=0$, so $x=-3$. That rules out (C) and (D).
The $-\ln$ reflects the graph in the $x$-axis, so it falls from left to right. That is (B). Check the $y$-intercept: $f(0)=-3-\ln 3\approx -4.1$.
Determine the equation of the asymptote of the function $f(x)=\log_9(x-3)-4$.
A log is undefined at 0, so the vertical asymptote is where the inside equals 0: $x-3=0$. The base and the $-4$ do not move it.
The function $f(x)$ has the form $f(x)=3\log_2(x+a)+b$. The function $g(x)$ has the form $g(x)=-\log_3(x+c)+5$. A section of the graphs of the two functions is shown.
Determine the values of $a$, $b$ and $c$.
Read the points you can trust off the grid. $f$ passes through $(0,1)$ and the intersection $(2,4)$. $g$ passes through $(0,5)$ and $(2,4)$.
Start with $g$, because it has one unknown. $g(0)=-\log_3 c+5=5$, so $\log_3 c=0$ and $c=1$. Check: $g(2)=-\log_3 3+5=4$.
For $f$, write one equation for each point:
$$3\log_2 a+b=1,\qquad 3\log_2(2+a)+b=4$$
Subtract them to get rid of $b$:
$$3\log_2\frac{2+a}{a}=3$$
$$\log_2\frac{2+a}{a}=1\ \Rightarrow\ \frac{2+a}{a}=2\ \Rightarrow\ a=2$$
Then $3\log_2 2+b=1$, so $b=-2$.
Consider the function $f(x)=\log_p(x+q)$ where $p>1$ and $0
The asymptote is $x=-q$, and with $0
Now the $y$-intercept: $f(0)=\log_p q$. With $p>1$ and $q<1$, that is negative. (A) goes through the origin, which would need $q=1$, and (D) has a positive $y$-intercept, which would need $q>1$.
Part 2
These used to be everywhere in early exams. Lots of technology-free papers opened with a couple of log equations you would recognise from Years 10 and 11. You still need the log laws, but these days they tend to sit inside a more complicated situation.
Use the inverse relationship to swap between forms:
$a^{n}=x\iff \log_a x=n$
Or equate the indices when the bases match:
$e^{2x}=e^{4}\ \Rightarrow\ 2x=4\ \Rightarrow\ x=2$
Watch the laws and the swap
Watch for equations that turn into a quadratic once you substitute. The QCAA have always loved those. And always check your answers in the original equation, because a log of a negative number does not exist.
Practise
Solve for $x$ in the equation $\ln(x)+\ln(2)=\ln(8)$.
Combine the left side with the first law: $\ln(2x)=\ln 8$, so $2x=8$. Option (D) is the trap: it adds instead of multiplies.
Solve $\ln(x)+\ln(3.70)=\ln(9.25)$ for $x$.
Same structure as the one above, with uglier numbers: $\ln(3.70x)=\ln 9.25$, so $x=\dfrac{9.25}{3.70}$. Option (D) subtracts instead of divides.
Solve for $x$ in the following.
(a) Swap to exponential form to get rid of the $\ln$: $2x=e^{5}$.
Divide by 2: $x=\dfrac{e^{5}}{2}$.
(b) Combine the logs with the second law:
$$\log_4\frac{4x+16}{x^{2}-2}=1$$
Swap forms and rearrange:
$$4x+16=4\left(x^{2}-2\right)\ \Rightarrow\ x^{2}-x-6=0$$
So $(x-3)(x+2)=0$, giving $x=3$ or $x=-2$.
There is no mark for checking here, but do it anyway. At $x=3$ the insides are 28 and 7. At $x=-2$ they are 8 and 2. Both are positive, so both solutions stand. That surprises a lot of students, who throw out the negative one out of habit.
Determine all possible solutions for the equation $\left(\log_2(x)\right)^{2}=\log_2\left(x^{2}\right)+8$. Evaluate the reasonableness of one solution.
It is not in base $e$, but it reduces to a quadratic, which the QCAA have always loved.
Use the third law on the right: $\log_2\left(x^{2}\right)=2\log_2 x$.
Now let $u=\log_2 x$:
$$u^{2}=2u+8\ \Rightarrow\ u^{2}-2u-8=0$$
$$(u-4)(u+2)=0$$
So $\log_2 x=4$ or $\log_2 x=-2$, giving $x=16$ or $x=\frac{1}{4}$.
Reasonableness, for $x=16$. Substitute into each side of the original equation. The left side is $\left(\log_2 16\right)^{2}=4^{2}=16$, and the right side is $\log_2 256+8=8+8=16$.
Both sides match, and $x$ and $x^{2}$ are positive so the logs exist. $x=16$ is reasonable.
Solve the following.
(a) Swap forms: $2x+4=5^{2}=25$, so $2x=21$.
(b) Combine the left side: $\ln\dfrac{7-x}{10}=\ln x$, so $7-x=10x$ and $11x=7$. Check: $7-\frac{7}{11}$ and $\frac{7}{11}$ are both positive.
Practise
Solve for $x$ in the equation $4+7e^{-2x}=3e^{2x}$.
This is a hidden quadratic. Multiply every term by $e^{2x}$ to clear the negative power, remembering $e^{-2x}\times e^{2x}=1$:
$$4e^{2x}+7=3e^{4x}$$
Let $u=e^{2x}$, so $e^{4x}=u^{2}$, and move everything to one side:
$$3u^{2}-4u-7=0$$
$$(3u-7)(u+1)=0$$
$u=-1$ is impossible, because $e^{2x}$ is always positive.
So $e^{2x}=\frac{7}{3}$, $2x=\ln\frac{7}{3}$ and $x=\frac{1}{2}\ln\frac{7}{3}$.
Solve the following equations.
(a) Divide by 4 first: $e^{x}=25$, so $x=\ln 25$.
(b) Use the third law, then the second, to get a single log:
$$\log_4\frac{x^{2}}{x-1}=1$$
Swap to exponential form and rearrange:
$$\frac{x^{2}}{x-1}=4\ \Rightarrow\ x^{2}-4x+4=0$$
So $(x-2)^{2}=0$ and $x=2$. Check: at $x=2$, both $x$ and $x-1$ are positive.
Part 3
These two rules are fundamental. You need to be completely fluent with them by the time you sit the external.
Watch the second rule
Same warning as with exponentials: being able to differentiate in this form is not enough on its own. Methods questions build through four phases, and you want to be confident at every one. Pick a phase to see the real questions that match it.
Straight use of the two rules. Simplify with the log laws first when you can. It often turns a chain rule into something much easier.
$\displaystyle\int_a^{5a}\frac{1}{x+a}\,dx,\ a\neq 0$ is
This one runs the rule backwards. Because $\frac{d}{dx}\ln(x+a)=\frac{1}{x+a}$, the antiderivative of $\frac{1}{x+a}$ is $\ln|x+a|$:
$$\Big[\ln|x+a|\Big]_a^{5a}=\ln|6a|-\ln|2a|=\ln 3\approx 1.0986$$
The $a$ cancels, whatever it is. On your calculator you can pick any value, say $a=1$, and get the same answer.
The slope of the tangent to the graph of $y=\ln\left(x^{2}\right)$ at $x=e^{2}$ is
Simplify first with the third law: $\ln\left(x^{2}\right)=2\ln x$, so $\dfrac{dy}{dx}=\dfrac{2}{x}$. Substitute $x=e^{2}$.
Determine $\dfrac{d}{dx}\big(3\ln(5x)\big)$.
Use $\frac{f'(x)}{f(x)}$ with $f(x)=5x$. The 5s cancel, which is worth noticing: the derivative of $\ln(kx)$ is always $\frac{1}{x}$.
$$3\times\frac{5}{5x}$$
Strictly speaking, this belongs to Unit 3 Topic 2. I have put it here so you can see the whole progression for logs in one place.
Differentiate $y=\ln(x)\cos(x)$ with respect to $x$.
Product rule with $u=\ln x$ and $v=\cos x$, so $u'=\frac{1}{x}$ and $v'=-\sin x$. The minus sign on $v'$ is what separates (C) from (D).
$$\frac{dy}{dx}=\frac{1}{x}\cos x+\ln x\,(-\sin x)$$
Given that $g(x)=\dfrac{\ln(x)}{x}$, determine the simplest value of $g'(e)$.
Quotient rule with $u=\ln x$ and $v=x$.
$$g'(x)=\frac{\frac{1}{x}\cdot x-\ln x\cdot 1}{x^{2}}=\frac{1-\ln x}{x^{2}}$$
At $x=e$, $\ln e=1$, so the top is $0$.
Worded does not mean hard. When the whole equation is given and you just need a derivative at a point, you can usually finish on your calculator. Read carefully, though: these questions love a distractor.
A football coach offered a 12-day intensive training clinic. During the clinic, the height that each player could kick a football was monitored. One player's kick heights could be modelled by $H(t)=\log_{10}(10t+10)+5,\ 0\le t\le 12$, where $H(t)$ is vertical height (m) and $t$ is the time (days) spent in training.
(a) Initial means $t=0$: $H(0)=\log_{10}10+5=6$ m.
(b) $\log_{10}(10t+10)=2$, so $10t+10=100$ and $t=9$ days.
(c) At the end of the course, $H(12)=\log_{10}130+5\approx 7.11$ m.
"Overall improvement" is a change, so subtract: $7.11-6\approx 1.11$ m. Giving $H(12)$ on its own is the common slip here.
(d) Differentiate on your calculator, or by hand:
$$H'(t)=\frac{1}{(t+1)\ln 10},\qquad H'(1.5)\approx 0.174\ \text{m/day}$$
(e) Solve $H'(t)=0.09$ on your calculator: $t\approx 3.83$ days.
The number of koalas in a conservation park is modelled by $N=15\ln(7t+1),\ t\ge 1$, where $t$ represents the time (years) since the park opened. There were 20 koalas in the park when it opened. Determine the approximate rate of change in the number of koalas when $t=3$.
"Rate of change" means differentiate. The 20 koalas is a distractor: you do not need it.
$$\frac{dN}{dt}=\frac{15\times 7}{7t+1}=\frac{105}{7t+1},\qquad N'(3)=\frac{105}{22}\approx 4.8$$
(A) is $N(3)$, the number of koalas rather than the rate.
Consider the function $f(x)=\ln(3x+4)$, for $x>\frac{-4}{3}$.
(a) $f'(x)=\dfrac{3}{3x+4}$.
(b) Let $f(x)=0$: $\ln(3x+4)=0$, so $3x+4=e^{0}=1$.
So $x=-1$.
(c) Substitute the intercept into the derivative: $f'(-1)=\dfrac{3}{1}=3$.
A new internet search engine gives a ranking $R$ to each website based on the function $R=\log_{10}\left(50h^{2}\right)$, where $h$ is the number of hits (visits) the website has received.
If a website currently has 100 hits, determine how many more hits they need to increase their ranking by 1.
No derivative here, even though it sits in a calculus context. Find the current ranking, add 1, and solve backwards.
Now: $R(100)=\log_{10}500\,000\approx 5.699$, so the new ranking is $6.699$.
$$\log_{10}\left(50h^{2}\right)=6.699\ \Rightarrow\ h\approx 316.2$$
Hits are whole numbers, so the site needs 317 hits, which is 217 more than it has.
A quick check: adding 1 to a $\log_{10}$ multiplies the inside by 10, so $h^{2}$ goes up by a factor of 10 and $h$ by $\sqrt{10}$.
The displacement of a particle (in metres) at time $t$ (in seconds) is represented by the function $s(t)=t\ln(t)-t,\ 0
Acceleration is the second derivative of displacement. The product rule on $t\ln t$ gives $\ln t+1$, and the $-t$ takes away the 1:
$$v(t)=\ln t,\qquad a(t)=\frac{1}{t},\qquad a(3)=\frac{1}{3}\approx 0.33$$
This is the one to aim for. The QCAA often give you a constant to find first, and you need it before you can go any further. It is a very common short response style, so make sure you are well practised before you sit your exam in November.
This continues the cockroach question from Topic 1.1. To decrease the growth rate of the cockroach population, a pest control treatment was trialled. The new function to model the cockroach population $t$ weeks after using the treatment is $I(t)=c\ln(t+8)+172$, where $c$ is a constant. Three weeks after using the treatment, the rate of population growth was five cockroaches per week.
(d) "Rate of population growth" is $I'(t)$, so differentiate before you substitute anything:
$$I'(t)=\frac{c}{t+8}$$
$$I'(3)=\frac{c}{11}=5\ \Rightarrow\ c=55$$
(e) "When the treatment began" is $t=0$: $I(0)=55\ln 8+172\approx 286.4$.
The magnitude of an earthquake can be modelled by the logarithmic equation $M_A=\log_{10}\left(\dfrac{I_A}{I_0}\right)$, where $M_A$ is the magnitude at a location A, $I_A$ is the intensity of the earthquake at location A and $I_0$ is a constant. An earthquake at location P had a magnitude of 5.2. A different earthquake at location Q had a magnitude of 3.5.
(a) Write one equation for each location and subtract:
$$M_P-M_Q=\log_{10}\frac{I_P}{I_0}-\log_{10}\frac{I_Q}{I_0}$$
(b) The second log law gets rid of $I_0$:
$$\log_{10}\frac{I_P}{I_Q}=M_P-M_Q$$
Swap to exponential form, so the intensities are the subject:
$$\frac{I_P}{I_Q}=10^{M_P-M_Q}$$
$$\frac{I_P}{I_Q}=10^{5.2-3.5}=10^{1.7}$$
So $\dfrac{I_P}{I_Q}\approx 50.1$.
Jaxon and Shari each own a shop and have recorded the number of customers entering their shop on two consecutive days.
The number of daily customers for each shop can be modelled by the equation $y=A\ln(Bx)$, where $x$ is the day and $y$ is the number of customers. The constants $A$ and $B$ are different for each shop.
Determine algebraically whether the total number of customers for Jaxon and Shari's shops will be the same every day in the future.
This one is complex unfamiliar, with its own full walkthrough.
Find $A$ and $B$ for each shop from its two days. The first law splits $A\ln(2B)$ into $A\ln 2+A\ln B$.
Shari: day 1 gives $A\ln B=10$ and day 2 gives $A\ln 2+A\ln B=20$. Subtracting, $A\ln 2=10$.
$$A=\frac{10}{\ln 2},\qquad \ln B=\frac{10}{A}=\ln 2\ \Rightarrow\ B=2$$
$$y_S=\frac{10}{\ln 2}\ln(2x)$$
Jaxon: $A\ln B=40$ and $A\ln 2+A\ln B=30$. Subtracting, $A\ln 2=-10$.
$$A=-\frac{10}{\ln 2},\qquad \ln B=\frac{40}{A}=-4\ln 2\ \Rightarrow\ B=2^{-4}=\frac{1}{16}$$
$$y_J=-\frac{10}{\ln 2}\ln\left(\frac{x}{16}\right)$$
Add the two models:
$$y_S+y_J=\frac{10}{\ln 2}\left[\ln(2x)-\ln\left(\frac{x}{16}\right)\right]$$
$$=\frac{10}{\ln 2}\ln 32=\frac{10\ln\left(2^{5}\right)}{\ln 2}=\frac{50\ln 2}{\ln 2}=50$$
The $x$ cancels inside the log, so the total does not depend on the day. It will be 50 every day in the future.
Game plan
Ask yourself these, in order, every time. They take you from reading the question to knowing exactly what to do. Try them on any Phase 3 or 4 question above.
The clue: if the answer needs a rate unit, like m/day or koalas per year, you differentiate.
Practice
Eight questions of my own, covering every skill. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. Try them with the game plan beside you.
State the equation of the asymptote and the exact coordinates of the $x$-intercept of $y=\ln(x+4)-2$.
$$\ln(x+4)=2\ \Rightarrow\ x+4=e^{2}\ \Rightarrow\ x=e^{2}-4$$
The question says exact, so leave it as $e^{2}-4$ rather than $3.39$.
The graph of $y=\ln x$ is the reflection of the graph of $y=e^{x}$ in
$y=e^{x}$ and $y=\ln x$ are inverses, and inverse functions are always reflections of each other in $y=x$. You can see it in the points: $(0,1)$ on $y=e^{x}$ swaps to $(1,0)$ on $y=\ln x$.
Solve $\ln x+\ln(x-3)=\ln 10$.
$$\ln\big(x(x-3)\big)=\ln 10$$
$$x^{2}-3x-10=0\ \Rightarrow\ (x-5)(x+2)=0$$
Solve $e^{2x}-5e^{x}+6=0$. Give your answers in exact form.
$$u^{2}-5u+6=0\ \Rightarrow\ (u-2)(u-3)=0$$
A population of bees is modelled by $P=200e^{0.15t}$, where $t$ is the time in weeks. Determine how long it takes for the population to reach 1000 bees.
$$e^{0.15t}=5$$
$$0.15t=\ln 5\ \Rightarrow\ t=\frac{\ln 5}{0.15}\approx 10.73$$
Determine $\dfrac{d}{dx}\ln\left(5x^{2}+1\right)$.
Given $f(x)=x^{2}\ln x$, determine the exact value of $f'(e)$.
$$u=x^{2}\quad u'=2x\qquad v=\ln x\quad v'=\frac{1}{x}$$
$$f'(x)=2x\ln x+x^{2}\cdot\frac{1}{x}=2x\ln x+x$$
$$f'(e)=2e(1)+e=3e$$
The number of wallabies in a sanctuary is modelled by $N=40\ln(3t+1)+25$, where $t$ is the time in years since the sanctuary opened. Determine the rate of change in the number of wallabies when $t=3$.
$$\frac{dN}{dt}=40\times\frac{3}{3t+1}=\frac{120}{3t+1}$$
$$\text{At } t=3:\quad \frac{dN}{dt}=\frac{120}{10}=12$$
Keep practising
Once Phase 4 feels comfortable, try the full walkthrough of 2023 Paper 1, Question 19. You will build two log models from a table, then show why their total never changes.
Past questions and marking-guide steps are from the 2020–2025 QCAA Mathematical Methods external assessments and sample assessment, © State of Queensland (QCAA) 2020–2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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