Logarithms Numerically and Graphically
Swap every $x$ and $y$ in a table of powers and you have a table of logarithms. Do the same to a graph and you get a reflection. The two functions are one relationship read from opposite ends.
The point $(3, 8)$ lies on $y = 2^x$. Without calculating anything, name a point that must lie on $y = \log_2 x$, and explain how you know. Then say what you think happens to the graph of $y = 2^x$ as $x$ becomes a large negative number, and whether the logarithm graph does anything similar.
$y = a^x$ and $y = \log_a x$ are inverse functions. One takes an index and returns a number; the other takes a number and returns the index. Swapping the roles of $x$ and $y$ turns each into the other, and on a graph swapping $x$ and $y$ is exactly a reflection in the line $y = x$.
$$(p, q) \text{ on } y = a^x \quad\Longleftrightarrow\quad (q, p) \text{ on } y = \log_a x$$
The exponential curve passes through $(0, 1)$ and hugs the $x$-axis on the left. The logarithm curve passes through $(1, 0)$ and hugs the $y$-axis below. Those two facts are the same fact, seen from either side of the mirror.
Know
- That $y = a^x$ and $y = \log_a x$ are inverses, and that their graphs reflect in $y = x$
- That $y = \log_a x$ has domain $x > 0$, range all real numbers, and a vertical asymptote at $x = 0$
- That $y = \log_a x$ is increasing when $a > 1$ and decreasing when $0 < a < 1$
Understand
- Why swapping $x$ and $y$ in a table is the same operation as reflecting in $y = x$
- Why the logarithm graph never touches the $y$-axis, however far down you follow it
Can Do
- Build a table of values for $y = \log_a x$ from one for $y = a^x$
- Sketch $y = \log_a x$ showing the intercept and the asymptote
- Decide from the base alone whether a logarithm graph rises or falls
Start with powers of $2$, which you already know, and read the same table both ways.
| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|---|---|
| $2^x$ | $\dfrac{1}{4}$ | $\dfrac{1}{2}$ | $1$ | $2$ | $4$ | $8$ |
Now read it from the bottom row up. The value $\dfrac{1}{4}$ came from the index $-2$, so $\log_2 \dfrac{1}{4} = -2$. The value $8$ came from the index $3$, so $\log_2 8 = 3$.
| $x$ | $\dfrac{1}{4}$ | $\dfrac{1}{2}$ | $1$ | $2$ | $4$ | $8$ |
|---|---|---|---|---|---|---|
| $\log_2 x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
Nothing was calculated. The second table is the first one with its rows swapped, which is what inverse means in practice.
Swapping the coordinates of every point is a reflection in the line $y = x$. Check it on a single point: $(3, 8)$ and $(8, 3)$ sit on opposite sides of that line, the same distance from it, on a segment perpendicular to it.
So every feature of the exponential graph has a partner on the logarithm graph:
| $y = 2^x$ | $y = \log_2 x$ |
|---|---|
| passes through $(0, 1)$ | passes through $(1, 0)$ |
| domain: all real $x$ | range: all real $y$ |
| range: $y > 0$ | domain: $x > 0$ |
| horizontal asymptote $y = 0$ | vertical asymptote $x = 0$ |
The right-hand column is not a new list to memorise. It is the left-hand column with the words $x$ and $y$ exchanged.
Three features are worth naming every time you sketch $y = \log_a x$ for $a > 1$.
It crosses the $x$-axis at $(1, 0)$. Any base raised to the power $0$ gives $1$, so $\log_a 1 = 0$ for every allowed base. All logarithm graphs pass through that one point.
It has a vertical asymptote at $x = 0$. As $x$ shrinks towards zero the logarithm plunges: $\log_2 \dfrac{1}{1024} = -10$, and there is no bottom to it. The curve never reaches the $y$-axis, because no index ever produces zero.
It grows, but slowly. To move the logarithm up by $1$ you must multiply $x$ by the base. Going from $\log_2 1024 = 10$ to $11$ needs $x = 2048$. Compare that with the exponential, which doubles its height with every single step right.
Everything so far assumed $a > 1$. A base between $0$ and $1$ is still allowed, and it turns the graph over.
Take $a = \dfrac{1}{2}$. Since $\left(\dfrac{1}{2}\right)^{3} = \dfrac{1}{8}$, we get $\log_{1/2} \dfrac{1}{8} = 3$, and since $\left(\dfrac{1}{2}\right)^{-3} = 8$, we get $\log_{1/2} 8 = -3$. Larger arguments now give smaller logarithms.
So the rule is decided by the base alone:
$$y = \log_a x \text{ is increasing when } a > 1, \text{ and decreasing when } 0 < a < 1.$$
Both versions still pass through $(1, 0)$ and still have the asymptote $x = 0$. Only the direction of travel changes.
Watch Me Solve It · 3 examples
-
1Write the given pairs$(-1, \tfrac{1}{3}), \; (0, 1), \; (1, 3), \; (2, 9)$These are the points on $y = 3^x$.
-
2Swap each coordinate pair$(\tfrac{1}{3}, -1), \; (1, 0), \; (3, 1), \; (9, 2)$The inverse sends each output back to its input.
-
3Read off the intercept$(1, 0)$The pair with $y = 0$ is where the curve meets the $x$-axis.
-
4Sanity check one value$\log_3 9 = 2 \iff 3^2 = 9 \;\checkmark$Translate back to index form; it should be a fact you already know.
-
1Start from the definition$y = \log_5 x \iff 5^y = x$Whatever $x$ is, it must be a power of $5$.
-
2Ask what values $5^y$ can take$5^y > 0 \text{ for every real } y$A positive base raised to any index is positive.
-
3Read off the domain$\text{domain: } x > 0$Only positive $x$ can be produced, so only positive $x$ is accepted.
-
4Read off the range and asymptote$\text{range: all real } y$$\text{asymptote: } x = 0$$y$ is unrestricted, and as $y \to -\infty$ the value $5^y$ approaches but never reaches $0$.
-
1Compare the base with 1$0 < 0.4 < 1$The base decides the direction on its own.
-
2Apply the rule$0 < a < 1 \Rightarrow y = \log_a x \text{ is decreasing}$
-
3Find the guaranteed point$\log_{0.4} 1 = 0 \Rightarrow (1, 0)$Every logarithm graph passes through $(1, 0)$.
-
4Find one more$0.4^1 = 0.4 \Rightarrow \log_{0.4} 0.4 = 1 \Rightarrow (0.4, 1)$The point $(a, 1)$ is always on $y = \log_a x$.
Brain Trainer · 4 problems
Four quick problems. Work each one, then reveal the answer.
-
1 The point $(2, 16)$ lies on $y = 4^x$. Name the matching point on $y = \log_4 x$.
Swap the pair.$(16, 2)$ -
2 State the domain and the asymptote of $y = \log_7 x$.
Only positive arguments are allowed, and the curve dives beside the $y$-axis.Domain $x > 0$, asymptote $x = 0$ -
3 Is $y = \log_{1/3} x$ increasing or decreasing?
The base $\dfrac{1}{3}$ lies between $0$ and $1$.Decreasing -
4 Every graph $y = \log_a x$ passes through which single point, whatever the base?
Because $a^0 = 1$ for every allowed base.$(1, 0)$
Multiple Choice · 5 questions
The graph of $y = \log_3 x$ is the graph of $y = 3^x$ reflected in:
The domain of $y = \log_6 x$ is:
The point $(3, 125)$ lies on $y = 5^x$. Which point must therefore lie on $y = \log_5 x$?
Which of these is a decreasing function?
As $x$ gets closer and closer to $0$ from the right, the graph of $y = \log_2 x$:
Short Answer · 3 questions
(a) Write the four corresponding points on $y = \log_4 x$.
(b) State the domain and range of $y = \log_4 x$.
(c) Explain why the two graphs meet the axes at different points.
(a) State one point that lies on both.
(b) Describe the relationship between the two curves.
(c) A student claims $\log_{1/2} 8 = 3$. Show that this is wrong and give the correct value.
(a) How far along the $x$-axis must you travel before the curve reaches a height of $6$?
(b) A page is $0.3$ m wide and the scale is $1$ cm per unit on both axes. Explain what goes wrong if you try to draw the curve up to $y = 6$ on that page.
(c) Suggest a change of scale that makes the sketch possible, and say what is lost.
Inverses
$(p, q)$ on $y = a^x$ gives $(q, p)$ on $y = \log_a x$
Mirror
Reflection in $y = x$
Always
Through $(1, 0)$, asymptote $x = 0$
Direction
Rises if $a > 1$, falls if $0 < a < 1$
Your Badges
0 of 6Mark lesson as complete
Tick when you've finished Learn, Practice and the Stretch. Earns +85 XP and +25 coins.