Graphs of Logarithmic Functions
Theory
The graph of \(y = \log_a x\) passes through \((1, 0)\), has the \(y\)-axis as a vertical asymptote and exists only for \(x \gt 0\). This Year 10A maths page covers increasing and decreasing log graphs, reflections and translations, intercepts, the link with \(y = a^x\), and finding an equation from a graph.
For a base \(a \gt 1\), the graph of \(y = \log_a x\) passes through \((1, 0)\) and \((a, 1)\). It is increasing, rising more and more slowly. As \(x\) gets close to \(0\), \(y\) heads down without limit, so the \(y\)-axis is a vertical asymptote.
The domain is \(x \gt 0\), because you can only take the log of a positive number. The range is all real numbers.
For a base between \(0\) and \(1\), such as \(\log_{0.5} x\), the graph still passes through \((1, 0)\) but is decreasing.
\(y = \log_a x\) and \(y = a^x\) undo each other. Their graphs are reflections in the line \(y = x\).
To find the asymptote and domain of any log graph, set the number inside the log equal to zero (asymptote) or greater than zero (domain).
This prepares you for Mathematical Methods and Specialist Mathematics in Years 11 and 12.
Transformations of \(y = \log_a x\) (with \(h, k \gt 0\)):
| Equation | Change to the graph | Asymptote | Domain |
|---|---|---|---|
| \(y = -\log_a x\) | reflect in the \(x\)-axis | \(x = 0\) | \(x \gt 0\) |
| \(y = \log_a(-x)\) | reflect in the \(y\)-axis | \(x = 0\) | \(x \lt 0\) |
| \(y = \log_a(x - h)\) | shift right \(h\) | \(x = h\) | \(x \gt h\) |
| \(y = \log_a(x + h)\) | shift left \(h\) | \(x = -h\) | \(x \gt -h\) |
| \(y = \log_a x + k\) | shift up \(k\) | \(x = 0\) | \(x \gt 0\) |
| \(y = \log_a(h - x)\) | reflect in \(y\)-axis, then shift right \(h\) | \(x = h\) | \(x \lt h\) |
How to sketch a log graph
- Asymptote: set the number inside the log equal to \(0\) and solve. Draw it as a dashed vertical line.
- Domain: the number inside the log is greater than \(0\). The graph lies on that side of the asymptote.
- Intercepts: let \(y = 0\) and solve in index form for the \(x\)-intercept; let \(x = 0\) for the \(y\)-intercept if \(0\) is in the domain.
- Direction: decide whether it rises or falls (base, minus signs), plot one more point, and draw the curve hugging the asymptote.
Matching and finding equations
- Choosing from options: use the asymptote first to cut the list down, then the direction, then test one marked point.
- Unknown base: substitute a point into \(y = \log_a x\) and change to index form, \(a^y = x\). The base must be positive.
- Two unknowns: the asymptote often gives one of them directly; otherwise substitute two points and solve the pair of equations (Example 3).
- Translations: left or right changes the \(x\) inside the log; up or down adds outside it.
- Intersections: set the two expressions equal. Write any whole number as a log of the same base, combine into one log on each side, then equate the numbers inside.
- How many meeting points: an increasing graph and a decreasing graph can cross at most once. Find one point that is on both and that is the only one.
Start from \(y = \log_2 x\). The \(+3\) inside moves it \(3\) left; the \(-1\) outside moves it \(1\) down.
Asymptote and domain. The number inside the log must be positive.
| \(x + 3\) | \(\gt\) | \(0\) |
| \(x\) | \(\gt\) | \(-3\) |
Asymptote \(\boldsymbol{x = -3}\), domain \(\boldsymbol{x \gt -3}\).
\(x\)-intercept. Let \(y = 0\).
| \(\log_2(x + 3)\) | \(=\) | \(1\) |
| \(x + 3\) | \(=\) | \(2^1\) |
| \(x\) | \(=\) | \(\mathbf{-1}\) |
\(y\)-intercept. Let \(x = 0\): \(y = \log_2 3 - 1 = 0.5849\ldots \approx \mathbf{0.58}\).
The graph exists only for \(x \lt 4\), so the number inside the log is positive exactly when \(x \lt 4\).
| \(k - x\) | \(\gt\) | \(0\) |
| \(x\) | \(\lt\) | \(k\) |
| \(k\) | \(=\) | \(\mathbf{4}\) |
Check each point in \(y = \log_2(4 - x)\):
| \(x = 3:\ \log_2 1\) | \(=\) | \(0\) |
| \(x = 0:\ \log_2 4\) | \(=\) | \(2\) |
| \(x = -4:\ \log_2 8\) | \(=\) | \(3\) |
The \(-x\) reflects \(y = \log_2 x\) in the \(y\)-axis, so the graph falls from left to right. A graph like this one is not \(y = -\log_2(x - 4)\): that graph lives to the right of \(x = 4\).
(i) The graph of \(y = \log_a x\) passes through \(\left(\dfrac{1}{8},\ -3\right)\). Find \(a\).
(ii) The graph of \(y = \log_a(x + b)\), where \(a \gt 1\), passes through \((1,\ 1)\) and \((13,\ 2)\). Find \(a\) and \(b\).
(i) Substitute the point and change to index form.
| \(-3\) | \(=\) | \(\log_a \dfrac{1}{8}\) |
| \(a^{-3}\) | \(=\) | \(\dfrac{1}{8} = 2^{-3}\) |
| \(a\) | \(=\) | \(\mathbf{2}\) |
(ii) Each point gives an equation in index form.
| \(1 = \log_a(1 + b)\) | \(\Rightarrow\) | \(a = 1 + b\) |
| \(2 = \log_a(13 + b)\) | \(\Rightarrow\) | \(a^2 = 13 + b\) |
Substitute \(b = a - 1\) into the second equation.
| \(a^2\) | \(=\) | \(13 + a - 1\) |
| \(a^2 - a - 12\) | \(=\) | \(0\) |
| \((a - 4)(a + 3)\) | \(=\) | \(0\) |
| \(a\) | \(=\) | \(\mathbf{4} \quad (a \gt 1)\) |
| \(b\) | \(=\) | \(4 - 1 = \mathbf{3}\) |
Check: \(\log_4 4 = 1\) and \(\log_4 16 = 2\).
(i) Find the point where \(y = \log_3 x\) and \(y = \log_3(x - 6) + 1\) intersect.
(ii) The graph of \(y = \log_3 x\) is reflected in the line \(y = x\). Write the equation of the image and the image of the intersection point.
(i) Write \(1\) as \(\log_3 3\) and combine the right-hand side into one log.
| \(\log_3 x\) | \(=\) | \(\log_3(x - 6) + \log_3 3\) |
| \(\log_3 x\) | \(=\) | \(\log_3\big(3(x - 6)\big)\) |
| \(x\) | \(=\) | \(3x - 18\) |
| \(x\) | \(=\) | \(9\) |
| \(y\) | \(=\) | \(\log_3 9 = 2\) |
The graphs meet at \(\mathbf{(9,\ 2)}\). Check: \(\log_3 3 + 1 = 2\).
(ii) Reflecting in \(y = x\) swaps \(x\) and \(y\).
| \(x\) | \(=\) | \(\log_3 y\) |
| \(y\) | \(=\) | \(\boldsymbol{3^x}\) |
The point \((9,\ 2)\) becomes \(\mathbf{(2,\ 9)}\). Check: \(3^2 = 9\).
Common pitfalls
Frequently asked questions
What does the graph of a log function look like?
For a base greater than \(1\), it passes through \((1, 0)\), rises slowly to the right and drops steeply towards the \(y\)-axis, which is a vertical asymptote. It only exists for \(x \gt 0\).
How do you find the asymptote of a log graph?
Set the number inside the log equal to zero and solve. For \(y = \log_2(x - 5)\) the asymptote is \(x = 5\). Adding or subtracting outside the log does not move it.
What is the domain of a logarithmic function?
The values of \(x\) that make the number inside the log positive. For \(y = \log_3(x + 4)\) the domain is \(x \gt -4\).
When is a log graph decreasing?
When the base is between \(0\) and \(1\), such as \(\log_{0.5} x\), or when there is a minus sign in front of a log with a base greater than \(1\).
How are log graphs and exponential graphs related?
\(y = \log_a x\) and \(y = a^x\) are inverse functions, so their graphs are reflections of each other in the line \(y = x\). A point \((p, q)\) on one becomes \((q, p)\) on the other.