Mathematical Methods · Units 3 & 4

Logarithmic Functions

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.

Learn

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…e^3 = 20.08\ldots. Reading it forwards you say ”ee to the power 33”. Reading it backwards a logarithm says “the power that turns ee into 20.08…20.08\ldots is 33”, which we write as log⁡e(20.08…)=3\log_e(20.08\ldots) = 3. The base of the log is the number being raised, and the answer of the log is the power.

So log⁡a(x)\log_a(x) is the power you raise aa to in order to get xx. 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 aanythinga^{\text{anything}} 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:

log⁡a(x)=y⟺ay=x\log_a(x) = y \quad \Longleftrightarrow \quad a^y = x

The star of Methods is the natural logarithm, written log⁡e(x)\log_e(x), whose base is the special number e≈2.718e \approx 2.718. When you see a log with no base shown in this course, treat it as base ee.

The shape of the graph and its wall

Picture the graph of y=log⁡e(x)y = \log_e(x). It climbs forever to the right, but slower and slower, like a runner who never quite stops. It crosses the xx axis at the point (1,0)(1, 0), because e0=1e^0 = 1 means the power that makes 11 is zero.

The interesting bit is on the left. As xx shrinks towards zero the graph plunges downward without limit, hugging the yy axis but never touching it. That invisible wall is the vertical asymptote, here the line x=0x = 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.

2468 -4-3-2-1123 (1, 0)
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=log⁡e(x)y = \log_e(x) is the inverse of y=exy = e^x. They undo each other, so log⁡e(ex)=x\log_e(e^x) = x and elog⁡e(x)=xe^{\log_e(x)} = x. On a graph, inverse functions are reflections of each other in the line y=xy = 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=0y = 0, so the log has a vertical asymptote at x=0x = 0. The exponential takes any xx and outputs only positives, so the log takes only positives and outputs any yy. Domain and range trade places. To find the rule of a log inverse, swap xx and yy in the exponential equation, then make yy the subject by taking log⁡e\log_e of both sides.

Moving and stretching the graph

Transformations of y=log⁡e(x)y = \log_e(x) follow the same rules as any function, but two features need watching because examiners target them every year.

  1. A change inside the bracket shifts the graph left or right and drags the vertical asymptote with it. For y=log⁡e(x−h)y = \log_e(x - h) the asymptote moves from x=0x = 0 to x=hx = h. The sign flips your instinct, so log⁡e(x−4)\log_e(x - 4) moves right and the asymptote lands at x=4x = 4.
  2. 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: log⁡a(x)=y  ⟺    ?\log_a(x) = y \iff \;?
Why must the input of a log be positive?
How do you find the vertical asymptote of y=log⁡e(x−4)+3y = \log_e(x - 4) + 3?
State the three log laws in one line.
What is the maximal domain of f(x)=log⁡e(5−x)f(x) = \log_e(5 - x)?
Why do some log-equation answers have to be rejected?
Recall · Exponential Functions
What is the horizontal asymptote of y=ex−4y = e^x - 4?
Recall · Inverse Functions
To find the inverse rule of f(x)=ex+2f(x) = e^x + 2, what is the first move?

Worked examples

Worked Example 1Solving a log equation

Solve log⁡e(2x−1)=3\log_e(2x - 1) = 3 for xx, giving an exact value.

  1. 1

    Rewrite the log statement as a power of ee. A log just asks what power you raise the base to.

    2x−1=e32x - 1 = e^3
  2. 2

    Solve the linear equation for xx.

    x=e3+12x = \frac{e^3 + 1}{2}
  3. 3

    Check the answer keeps the argument positive, since log⁡e\log_e only accepts inputs greater than zero. Here 2x−1=e3>02x - 1 = e^3 > 0, so it is valid.

    2x−1=e3>0✓2x - 1 = e^3 > 0 \quad \checkmark
Answer
x=e3+12x = \dfrac{e^3 + 1}{2}
Worked Example 2A log equation with a rejected solution

Solve log⁡e(x)+log⁡e(x−3)=log⁡e(4)\log_e(x) + \log_e(x - 3) = \log_e(4) for xx.

  1. 1

    Combine the two logs on the left using log⁡e(a)+log⁡e(b)=log⁡e(ab)\log_e(a) + \log_e(b) = \log_e(ab).

    log⁡e(x(x−3))=log⁡e(4)\log_e(x(x - 3)) = \log_e(4)
  2. 2

    Equal logs with the same base mean equal arguments. Drop the logs.

    x(x−3)=4x(x - 3) = 4
  3. 3

    Expand and solve the quadratic.

    x2−3x−4=0  ⟹  (x−4)(x+1)=0x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0
  4. 4

    Both x=4x = 4 and x=−1x = -1 appear, but the original log⁡e(x)\log_e(x) needs x>0x > 0. Reject x=−1x = -1.

    x=4(reject x=−1)x = 4 \quad (\text{reject } x = -1)
Answer
x=4x = 4
Worked Example 3Transforming the natural log graph

The graph of y=log⁡e(x)y = \log_e(x) is transformed to y=log⁡e(x+2)−1y = \log_e(x + 2) - 1. State the equation of the vertical asymptote and the exact xx intercept of the new graph.

  1. 1

    The +2+2 inside the bracket is a translation 22 units to the left. The asymptote of y=log⁡e(x)y = \log_e(x) sits at x=0x = 0, so it moves with the graph.

    x=−2x = -2
  2. 2

    For the xx intercept set y=0y = 0 and rewrite as a power of ee.

    log⁡e(x+2)−1=0  ⟹  log⁡e(x+2)=1\log_e(x + 2) - 1 = 0 \implies \log_e(x + 2) = 1
  3. 3

    Undo the log. The argument equals e1e^1.

    x+2=e  ⟹  x=e−2x + 2 = e \implies x = e - 2
Answer
Asymptotex=−2,andxinterceptatx=e−2Asymptote x = -2, and x intercept at x = 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=log⁡e(x+1)−3y = \log_e(x + 1) - 3 are

1mark

VCAA 2024 Mathematical Methods Exam 2, Section A Q1

Q2.The equation of the vertical asymptote of the graph of y=log⁡e(x−4)+3y = \log_e(x - 4) + 3 is:

1mark
Need a hint?
The asymptote moves with whatever is inside the bracket. Set x−4=0x - 4 = 0 and remember the +3+3 only shifts the graph up and down.

Q3.The maximal domain of the function f(x)=log⁡e(5−x)f(x) = \log_e(5 - x) is:

1mark
Need a hint?
The argument of a log must be strictly positive. Set 5−x>05 - x > 0 and solve the inequality.

Q4.The inverse function of f(x)=ex+2f(x) = e^{x} + 2, with domain R\mathbb{R}, is:

1mark
Need a hint?
Swap xx and yy, then make yy the subject. Undo the +2+2 before you take the log, so the 22 stays inside the argument.

Q5.The solution to 2log⁡e(x)=log⁡e(2x+3)2\log_e(x) = \log_e(2x + 3) is:

1mark
Need a hint?
Use the power law to turn 2log⁡e(x)2\log_e(x) into a single log, drop the logs, then check each solution keeps x>0x > 0.

Q6.The exact xx intercept of the graph of y=log⁡e(3x)−2y = \log_e(3x) - 2 is:

1mark
Need a hint?
Set y=0y = 0 to get log⁡e(3x)=2\log_e(3x) = 2, then rewrite as a power of ee to undo the log before solving for xx.

Q7.Solve log⁡e(x+5)+log⁡e(x+1)=log⁡e(5)\log_e(x + 5) + \log_e(x + 1) = \log_e(5) for xx. Show every step and state any rejected solution.

3marks

Work this on paper. The worked solution appears once you submit.

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.