How to Graph Logarithmic Functions?
TL;DR: A logarithmic graph is the inverse of an exponential graph. The parent graph \(y=\log_b(x)\) has domain \(x>0\), range all real numbers, vertical asymptote \(x=0\), and key points \((1,0)\), \((b,1)\), and \((b^2,2)\).
- If \(b>1\), the parent log graph increases. If \(0
- For \(y=a\log_b(x-h)+k\), the vertical asymptote moves to \(x=h\).
- Use easy log inputs \(1\), \(b\), and \(b^2\) to build reliable points quickly.
- A log graph often has no y-intercept because the input of the logarithm must be positive.
Graphing logarithmic functions is much easier when you stop hunting for random x-values. A log graph has a wall, called a vertical asymptote, and it has a few dependable points that come from powers of the base. In this lesson, we will use those pieces like a tutor would: find the asymptote, build easy points, apply the transformations, and check the shape.
Graph Logarithmic Functions: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
A reliable path
- Translate firstAsk: the base to what power gives the input?
- Use rules legallyProducts, quotients, and powers have rules; sums do not split.
- Protect the domainKeep the log input positive and track asymptotes when graphing.
Worked examples
Evaluate a log
- Ask 3 to what power equals 81.
- 3 to the fourth power is 81.
- The logarithm is that exponent.
Find a log domain
- The input is x – 5.
- Require x – 5 > 0.
- Solve the inequality.
Try one before moving on
Graph Logarithmic Functions: pop-up practice
What a logarithmic graph means
A logarithm asks an exponent question:
So \(\log_2(8)=3\) because \(2^3=8\). On a graph, the input \(x\) must be positive, because there is no real exponent that makes a positive base equal to zero or a negative number.
The parent graph \(y=\log_b(x)\)
| Feature | What it means |
|---|---|
| Base | \(b>0\) and \(b\ne1\). Bases greater than 1 increase; bases between 0 and 1 decrease. |
| Domain | \(x>0\), because the log input must be positive. |
| Range | All real numbers. The graph can go up or down without a highest or lowest y-value. |
| Vertical asymptote | \(x=0\). The graph approaches the y-axis but never touches it. |
| Key points | \((1,0)\), \((b,1)\), and \((b^2,2)\). These come from \(\log_b(1)=0\), \(\log_b(b)=1\), and \(\log_b(b^2)=2\). |
| Inverse relationship | \(y=\log_b(x)\) reflects \(y=b^x\) across \(y=x\). |
| One-to-one | Every horizontal line hits the graph at most once, so a logarithmic function has an inverse. |
End behavior and special log facts
The graph behavior near the asymptote depends on the base.
| Base range | As \(x\to0^+\) | As \(x\to\infty\) | Shape |
|---|---|---|---|
| \(b>1\) | \(y\to-\infty\) | \(y\to\infty\) | Increasing |
\(0| \(y\to\infty\) | \(y\to-\infty\) | Decreasing | |
Two more facts are worth keeping close:
- \(\ln(x)=\log_e(x)\), where \(e\approx2.718\).
- Change of base: \(\log_b(x)=\dfrac{\ln x}{\ln b}\). This is why graphing tools can use natural logs to graph logs with many different bases.
How to graph \(y=a\log_b(x-h)+k\)
Most log graph problems in school math are transformations of the parent graph. Use this reliable process:
- Find the vertical asymptote. Set the log input equal to zero. For \(x-h\), the asymptote is \(x=h\).
- Find the domain. Make the log input greater than zero. For \(x-h\), the domain is \(x>h\).
- Choose easy inputs. Make the inside of the log equal \(1\), \(b\), and \(b^2\).
- Transform the y-values. The parent y-values \(0\), \(1\), and \(2\) become \(k\), \(a+k\), and \(2a+k\).
- Sketch the curve. The curve should approach the asymptote and pass through your transformed points.
Those three points are the fastest clean graphing points for \(y=a\log_b(x-h)+k\).
Worked example 1: graph \(f(x)=2\log_3(x+1)\)
Step-by-step
- The input is \(x+1\), so set \(x+1=0\). The vertical asymptote is \(x=-1\).
- The domain is \(x+1>0\), so \(x>-1\).
- Use inside values \(1\), \(3\), and \(9\). That gives x-values \(0\), \(2\), and \(8\).
- The parent outputs \(0\), \(1\), and \(2\) are multiplied by 2, so the y-values are \(0\), \(2\), and \(4\).
Graphing data
Asymptote: \(x=-1\)
Domain: \((-1,\infty)\)
Range: all real numbers
Points: \((0,0)\), \((2,2)\), \((8,4)\)
The graph is increasing because the base is \(3>1\) and the coefficient is positive.
Worked example 2: graph \(g(x)=-\log_2(x-4)+1\)
This one has a horizontal shift, a reflection across a horizontal line, and a vertical shift.
- Set \(x-4=0\). The vertical asymptote is \(x=4\), and the domain is \(x>4\).
- Use inside values \(1\), \(2\), and \(4\), giving x-values \(5\), \(6\), and \(8\).
- The parent outputs are \(0\), \(1\), and \(2\).
- Multiply by \(-1\) and add \(1\), giving y-values \(1\), \(0\), and \(-1\).
Graph \((5,1)\), \((6,0)\), and \((8,-1)\), then sketch a decreasing curve approaching \(x=4\).
Worked example 3: graph \(h(x)=\log_{1/2}(x)+2\)
The base is between 0 and 1, so the graph decreases as x increases.
| Inside value | Log value | After adding 2 | Point |
|---|---|---|---|
| \(1\) | \(0\) | \(2\) | \((1,2)\) |
| \(1/2\) | \(1\) | \(3\) | \((1/2,3)\) |
| \(2\) | \(-1\) | \(1\) | \((2,1)\) |
Common mistakes to avoid
- Using random x-values first. Pick inside values \(1\), \(b\), and \(b^2\) instead. They make the log values easy.
- Forgetting the domain. The inside of the log must be greater than zero, not greater than or equal to zero.
- Drawing a y-intercept automatically. Many log graphs never touch the y-axis.
- Mixing up shifts. In \(\log_b(x-h)\), the asymptote is \(x=h\), not \(x=-h\).
- Ignoring a negative coefficient. A negative value of \(a\) flips the graph vertically.
Practice problems
- For \(f(x)=\log_4(x-2)\), find the vertical asymptote and domain.
- For \(g(x)=3\log_2(x)+1\), list three easy points.
- For \(h(x)=-2\log_5(x+3)\), find the asymptote and say whether the graph increases or decreases.
- For \(p(x)=\log_{1/3}(x)-4\), find the point that comes from the parent point \((1,0)\).
- For \(q(x)=2\log_3(x-1)-5\), find the asymptote, domain, range, three graphing points, and whether the graph increases or decreases.
Answers:
- Asymptote \(x=2\); domain \(x>2\).
- Use inside values \(1,2,4\): points \((1,1)\), \((2,4)\), \((4,7)\).
- Asymptote \(x=-3\). The graph decreases because the base is greater than 1 and the coefficient is negative.
- The point \((1,0)\) shifts down 4, so \((1,-4)\).
- Asymptote \(x=1\); domain \(x>1\); range all real numbers. Use inside values \(1,3,9\), so x-values are \(2,4,10\). Parent outputs \(0,1,2\) become \(-5,-3,-1\). Points: \((2,-5)\), \((4,-3)\), \((10,-1)\). The graph increases because \(3>1\) and the coefficient is positive.
Related EffortlessMath lessons
If a piece of this lesson feels rusty, these pages connect directly to the skills used here:
Recommended EffortlessMath Books
For a workbook that builds logarithmic graphs into a full exponential-and-log unit, the Algebra II for Beginners walks through every transformation with worked examples. For pre-calc-level coverage that prepares you for calculus and natural logs, see the Pre-Calculus for Beginners.
Frequently Asked Questions
How do you graph a logarithmic function?
Find the vertical asymptote by setting the log input equal to zero, choose easy inputs such as 1, the base, and the base squared, transform those points, and sketch a smooth curve that approaches the asymptote.
What are the domain and range of a logarithmic function?
The log input must be positive. For \(y=\log_b(x)\), the domain is \(x>0\) and the range is all real numbers. For \(y=\log_b(x-h)+k\), solve \(x-h>0\), so the domain is \(x>h\).
What is the vertical asymptote of a log graph?
For the parent function \(y=\log_b(x)\), the vertical asymptote is \(x=0\). For \(y=a\log_b(x-h)+k\), the asymptote is \(x=h\).
Does every logarithmic function have an x-intercept?
Most transformed log functions have an x-intercept, but its location depends on the vertical shift. Set \(y=0\) and solve. For \(y=\log_b(x)\), the x-intercept is always \((1,0)\).
Does a logarithmic function have a y-intercept?
Only sometimes. A y-intercept exists only if \(x=0\) is in the domain. For \(y=\log_b(x)\), there is no y-intercept because \(\log_b(0)\) is undefined.
How are logarithmic and exponential graphs related?
They are inverse functions. The graph of \(y=\log_b(x)\) is the reflection of \(y=b^x\) across the line \(y=x\).
What happens when the log base is between 0 and 1?
When \(0
Why is a logarithmic function one-to-one?
A logarithmic function passes the horizontal line test. For each output, there is exactly one input, which is why logs have inverse exponential functions.
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