Master logarithmic functions the easy way, with plain English intuition, the natural log, vertical asymptotes, the inverse link to exponentials, transformations, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.
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Press a button on your phone and you raise a number to a power. Logarithms are the
undo button. They run the whole thing backwards and ask the only question that matters:
“what power was used to get here?” That is the entire idea. A logarithm is a power
detective. Once you see it that way, every graph, asymptote and equation in this topic
becomes a story about powers that have been hidden and need finding.
What a logarithm actually asks
Start with a power statement you already trust, like e3=20.08…. Reading it
forwards you say ”e to the power 3”. Reading it backwards a logarithm says
“the power that turns e into 20.08… is 3”, which we write as
loge(20.08…)=3. The base of the log is the number being raised, and the
answer of the log is the power.
So loga(x) is the power you raise a to in order to get x. Two facts fall
straight out of that sentence and they fix most mistakes before they start. First, you
can only ask for the power that produces a positive number, because aanything
is always positive. That is why the input of every log must be greater than zero.
Second, a log and its matching power statement are two ways of saying the same thing:
loga(x)=y⟺ay=x
The star of Methods is the natural logarithm, written loge(x), whose base is
the special number e≈2.718. When you see a log with no base shown in this
course, treat it as base e.
The shape of the graph and its wall
Picture the graph of y=loge(x). It climbs forever to the right, but slower and
slower, like a runner who never quite stops. It crosses the x axis at the point
(1,0), because e0=1 means the power that makes 1 is zero.
The interesting bit is on the left. As x shrinks towards zero the graph plunges
downward without limit, hugging the y axis but never touching it. That invisible
wall is the vertical asymptote, here the line x=0. It appears because the input
of a log can never reach zero, so the curve has nowhere to land. Every basic log graph
has exactly one vertical asymptote, and finding where it sits is worth easy marks.
The graph of y = log_e(x) crosses the x axis at (1, 0) and plunges down against its vertical asymptote, the line x = 0.
Logs and exponentials are mirror twins
Here is the cleanest way to understand logs. The function y=loge(x) is the
inverse of y=ex. They undo each other, so loge(ex)=x and
eloge(x)=x. On a graph, inverse functions are reflections of each other in the
line y=x, which is why the log graph is just the exponential graph flipped across
that diagonal.
That reflection swaps everything. The exponential has a horizontal asymptote at
y=0, so the log has a vertical asymptote at x=0. The exponential takes any
x and outputs only positives, so the log takes only positives and outputs any y.
Domain and range trade places. To find the rule of a log inverse, swap x and y in
the exponential equation, then make y the subject by taking loge of both sides.
Moving and stretching the graph
Transformations of y=loge(x) follow the same rules as any function, but two
features need watching because examiners target them every year.
A change inside the bracket shifts the graph left or right and drags the vertical
asymptote with it. For y=loge(x−h) the asymptote moves from x=0 to
x=h. The sign flips your instinct, so loge(x−4) moves right and the
asymptote lands at x=4.
A change outside the log, added or multiplied, shifts or stretches the graph up
and down. This never moves a vertical asymptote, but students forget the vertical
translation constantly, so write it down.
To solve a log equation, get a single log by itself, rewrite it as a power of the base
to peel the log away, then solve what is left. Finish by checking every answer against
the domain. Reject any value that makes an argument zero or negative, because it was
never a real solution, only a ghost created by the algebra.
See these moves in action in the Worked Examples tab, then test yourself in
Try It.
Lock it in with active recall
Active recall
Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).
Complete the equivalence: loga(x)=y⟺?
ay=x. A log and its matching power statement say exactly the same thing.
Why must the input of a log be positive?
A log asks what power makes the input, and a positive base raised to any power is always positive — it can never produce zero or a negative.
How do you find the vertical asymptote of y=loge(x−4)+3?
Set the argument to zero: x−4=0, so x=4. The outside +3 does not move a vertical asymptote.
State the three log laws in one line.
log(xy)=logx+logy, logyx=logx−logy, and log(xn)=nlogx — product to sum, quotient to difference, power to the front.
What is the maximal domain of f(x)=loge(5−x)?
(−∞,5). The argument must be strictly positive: 5−x>0⇒x<5.
Why do some log-equation answers have to be rejected?
The algebra can hand you a value that makes an original log take a zero or negative input. Substitute back and throw away any that break the domain.
Recall · Exponential Functions
What is the horizontal asymptote of y=ex−4?
y=−4. Subtracting 4 slides the whole curve, and its floor, down by 4.
Recall · Inverse Functions
To find the inverse rule of f(x)=ex+2, what is the first move?
Swap x and y: x=ey+2, then make y the subject, giving f−1(x)=loge(x−2).
Worked examples
Worked Example 1Solving a log equation
Solve loge(2x−1)=3 for x, giving an exact value.
1
Rewrite the log statement as a power of e. A log just asks what power you raise the base to.
2x−1=e3
2
Solve the linear equation for x.
x=2e3+1
3
Check the answer keeps the argument positive, since loge only accepts inputs greater than zero. Here 2x−1=e3>0, so it is valid.
2x−1=e3>0✓
Answer
x=2e3+1
Worked Example 2A log equation with a rejected solution
Solve loge(x)+loge(x−3)=loge(4) for x.
1
Combine the two logs on the left using loge(a)+loge(b)=loge(ab).
loge(x(x−3))=loge(4)
2
Equal logs with the same base mean equal arguments. Drop the logs.
x(x−3)=4
3
Expand and solve the quadratic.
x2−3x−4=0⟹(x−4)(x+1)=0
4
Both x=4 and x=−1 appear, but the original loge(x) needs x>0. Reject x=−1.
x=4(reject x=−1)
Answer
x=4
Worked Example 3Transforming the natural log graph
The graph of y=loge(x) is transformed to y=loge(x+2)−1. State the equation of the vertical asymptote and the exact x intercept of the new graph.
1
The +2 inside the bracket is a translation 2 units to the left. The asymptote of y=loge(x) sits at x=0, so it moves with the graph.
x=−2
2
For the x intercept set y=0 and rewrite as a power of e.
loge(x+2)−1=0⟹loge(x+2)=1
3
Undo the log. The argument equals e1.
x+2=e⟹x=e−2
Answer
Asymptotex=−2,andxinterceptatx=e−2
Practice questions
Practice test
Try it yourself
7 questions, 9 marks
Choose your answers, then submit to see your score and the full worked solutions.
Multiple choice is marked for you, just like Exam 2 Section A.
Q1.The asymptote(s) of the graph of y=loge(x+1)−3 are
1mark
Show worked solution
A logarithmic function loge(x+1) has a single vertical asymptote at x=−1 where the argument is zero. The −3 is a vertical shift and creates no horizontal asymptote, so the only asymptote is x=−1.
VCAA 2024 Mathematical Methods Exam 2, Section A Q1
Q2.The equation of the vertical asymptote of the graph of y=loge(x−4)+3 is:
1mark
Need a hint?
The asymptote moves with whatever is inside the bracket. Set x−4=0 and remember the +3 only shifts the graph up and down.
Show worked solution
The asymptote of y=loge(x) is the line x=0. The −4 inside the bracket translates the graph 4 units to the right, carrying the asymptote to x=4. The +3 is a vertical translation and does not move a vertical asymptote. Option D, x=−4, is the classic sign slip, moving the graph the wrong way.
Q3.The maximal domain of the function f(x)=loge(5−x) is:
1mark
Need a hint?
The argument of a log must be strictly positive. Set 5−x>0 and solve the inequality.
Show worked solution
A log only accepts a strictly positive argument, so we need 5−x>0, which rearranges to x<5. In interval notation that is (−∞,5). Option C reverses the inequality, and option B wrongly includes the endpoint with a square bracket even though loge(0) is undefined.
Q4.The inverse function of f(x)=ex+2, with domain R, is:
1mark
Need a hint?
Swap x and y, then make y the subject. Undo the +2 before you take the log, so the 2 stays inside the argument.
Show worked solution
Swap x and y, so x=ey+2. Subtract 2 to get x−2=ey, then take the natural log of both sides: y=loge(x−2). Options A and D subtract or add the 2 outside the log, a common error from applying the inverse to the wrong part. The −2 must stay inside the argument.
Q5.The solution to 2loge(x)=loge(2x+3) is:
1mark
Need a hint?
Use the power law to turn 2loge(x) into a single log, drop the logs, then check each solution keeps x>0.
Show worked solution
Use the power law: 2loge(x)=loge(x2), so x2=2x+3, giving x2−2x−3=0 and (x−3)(x+1)=0. Both x=3 and x=−1 solve the quadratic, but loge(x) requires x>0, so x=−1 must be rejected. Option C keeps the invalid solution, the single most common slip in log equations.
Q6.The exact x intercept of the graph of y=loge(3x)−2 is:
1mark
Need a hint?
Set y=0 to get loge(3x)=2, then rewrite as a power of e to undo the log before solving for x.
Show worked solution
Set y=0, so loge(3x)=2. Undo the log by writing the argument as e2: 3x=e2, hence x=3e2. Option A comes from treating loge(3x)=2 as 3x=2, forgetting to undo the log with e.
Q7.Solve loge(x+5)+loge(x+1)=loge(5) for x. Show every step and state any rejected solution.
3marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
Combine the logs on the left with the addition law:
loge((x+5)(x+1))=loge(5).
Equal logs of the same base give equal arguments:
(x+5)(x+1)=5.
Expand and simplify:
x2+6x+5=5⟹x2+6x=0⟹x(x+6)=0.
So x=0 or x=−6. The original equation needs x+5>0 and x+1>0, that is x>−1. Therefore x=−6 is rejected and the only solution is
x=0.
Frequently asked questions
Why must the input of a log be positive?
A logarithm asks what power you raise the base to in order to get the input. Since a positive base raised to any power is always positive, you can never produce zero or a negative number, so the input of a log has to be greater than zero.
How do I find the vertical asymptote of a log graph?
Set whatever is inside the log equal to zero and solve for x. That x value is the vertical asymptote, because the log shoots off to negative infinity as its input approaches zero. For example, the asymptote of log base e of (x minus 4) is the line x equals 4.
Why do some log equation answers have to be rejected?
When you combine logs and solve the resulting quadratic, the algebra can hand you a value that makes one of the original logs take a zero or negative input. That value was never a genuine solution, so always substitute each answer back and throw away any that break the domain.