Graphs Of Logarithm Functions
Understand the graphs of logarithm functions for Queensland Year 11 Mathematical Methods (QCAA). A logarithm graph rises slowly and is the reflection of an exponential graph in the line y equals x.
You will learn to sketch the curve, locate its vertical asymptote and intercept, state the domain and range, and describe the effect of the parameters that shift the graph — with applications such as decibels and the Richter scale.
Every question with a fully worked solution.
- Graphs Of Logarithm Functions - Video - Graphs of logarithmic functions Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), the graph of \(y=\log_a x\) (with \(a>1\)) is an increasing curve that rises slowly, passes through \((1,0)\), and has a vertical asymptote at \(x=0\). This page describes those qualitative features, the effect of the parameters \(h\) and \(k\) in \(y=\log_a(x-h)+k\), how to find the \(x\)-intercept, and the reflection of \(y=a^{x}\) in the line \(y=x\).
The logarithm function \(y=\log_a x\) (for a base \(a>1\)) is the inverse of the exponential \(y=a^{x}\). It is defined only for \(x>0\), is always increasing, and grows more and more slowly as \(x\) increases. Every such graph passes through \((1,0)\) because \(\log_a 1=0\).
The line \(x=0\) is a vertical asymptote: as \(x\) approaches \(0\) from the right the curve plunges downwards without ever touching the \(y\)-axis. For the transformed curve \(y=\log_a(x-h)+k\), the parameter \(h\) shifts everything right, so the asymptote becomes \(x=h\), the domain is \(x>h\), and the parameter \(k\) shifts the curve up. The range is always all real numbers.
Because \(y=\log_a x\) and \(y=a^{x}\) are inverses, each graph is the reflection of the other in the line \(y=x\).
General transformed logarithm function (\(a>1\)):
Vertical asymptote and domain:
Passes through the point where the log is zero, i.e. \(x-h=1\):
How to sketch or analyse \(y=\log_a(x-h)+k\)
- Asymptote: set the argument to zero, \(x-h=0\), so the vertical asymptote is \(x=h\) and the domain is \(x>h\).
- Key point: the log is zero when \(x-h=1\), giving the point \((h+1,\,k)\); plot one or two more values to show the slow rise.
- \(x\)-intercept: set \(y=0\), isolate the logarithm, then rewrite in index form to solve for \(x\).
Asymptote — the argument is \(x\), so set it to zero:
| \(x\) | \(=\) | \(0\) |
The vertical asymptote is the line \(x=0\) (the \(y\)-axis), so the domain is \(x>0\).
A point — the log is zero when the argument is \(1\):
| \(\log_2 1\) | \(=\) | \(0\) |
So the curve passes through \((1,0)\); it is increasing and rises slowly (e.g. \((2,1)\), \((4,2)\)).
Asymptote \(x=0\), domain \(x>0\), through \((1,0)\); increasing and slow.
Set \(y=0\) and isolate the logarithm:
| \(0\) | \(=\) | \(\log_3 x-2\) |
| \(\log_3 x\) | \(=\) | \(2\) |
Rewrite in index form and equate:
| \(x\) | \(=\) | \(3^{2}\) |
| \(x\) | \(=\) | \(9\) |
\(x\)-intercept at \((9,\,0)\) (asymptote still \(x=0\)).
Asymptote — set the argument to zero:
| \(x-1\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(1\) |
Vertical asymptote \(x=1\), so the domain is \(x>1\).
\(x\)-intercept — set \(y=0\), then use index form:
| \(\log_2(x-1)\) | \(=\) | \(0\) |
| \(x-1\) | \(=\) | \(2^{0}=1\) |
| \(x\) | \(=\) | \(2\) |
Asymptote \(x=1\), domain \(x>1\), \(x\)-intercept \((2,\,0)\).
Asymptote — set the argument to zero (here \(h=-4\)):
| \(x+4\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-4\) |
Vertical asymptote \(x=-4\), so the domain is \(x>-4\).
\(x\)-intercept — set \(y=0\), isolate the log, then index form:
| \(0\) | \(=\) | \(\log_5(x+4)-1\) |
| \(\log_5(x+4)\) | \(=\) | \(1\) |
| \(x+4\) | \(=\) | \(5^{1}=5\) |
| \(x\) | \(=\) | \(1\) |
Asymptote \(x=-4\), domain \(x>-4\), \(x\)-intercept \((1,\,0)\).
Common pitfalls
Frequently asked questions
What does the graph of y = log_a x look like for a greater than 1?
An increasing curve that rises slowly, passes through \((1,0)\), and has a vertical asymptote at \(x=0\); the domain is \(x>0\).
Where is the vertical asymptote of y = log_a(x - h) + k?
At \(x=h\) — the value of \(x\) that makes the argument \(x-h\) equal to zero. The domain is \(x>h\).
How do you find the x-intercept of a logarithm graph?
Set \(y=0\), isolate the logarithm, then rewrite it in index form and solve for \(x\).
What is the domain and range of y = log_a x?
The domain is \(x>0\) and the range is all real numbers, because the curve keeps rising without bound.
How is the log graph related to the exponential graph?
\(y=\log_a x\) is the inverse of \(y=a^{x}\), so it is the reflection of the exponential curve in the line \(y=x\).
Does a logarithm graph ever reach or cross its asymptote?
No. The curve gets closer and closer to the vertical asymptote \(x=h\) but never touches or crosses it.