Logarithmic Graphs

Start from $y=2^x$y = 2ˣ

The graph of $y=2^x$y = 2ˣ passes through (0, 1), (1, 2), (2, 4) and (3, 8), doubling its height at each step to the right. To the left it halves instead, through $\left(-1,\frac{1}{2}\right)$(−1, 1/2) and $\left(-2,\frac{1}{4}\right)$(−2, 1/4), closing in on the x-axis without reaching it. Every power of 2 is positive, so the whole curve lies above the x-axis.

So $y=2^x$y = 2ˣ takes any number x as an input, and its outputs are the positive numbers only. The logarithm to base 2 undoes it: ${\log }_{2}x$log₂ x is the power of 2 that gives x. Its graph is found from the graph of $y=2^x$y = 2ˣ in one move.

Swapping x and y is a reflection

A function and its inverse undo each other. If $y=2^x$y = 2ˣ sends 2 to 4, then ${\log }_2$log₂ sends 4 back to 2. So the point (2, 4) on the graph of $y=2^x$y = 2ˣ becomes the point (4, 2) on the graph of $y={\log }_{2}x$y = log₂ x. The input and the output change places.

Swapping the coordinates of a point reflects it in the line y = x. Take (2, 4) and (4, 2). The segment joining them has gradient (2 − 4) ÷ (4 − 2) = −1, and the line y = x has gradient 1. The product of the gradients is −1, so the segment crosses y = x at a right angle. Its midpoint is $\left(\frac{\left(2+4\right)}{2},\frac{\left(4+2\right)}{2}\right)=\left(3,3\right)$((2 + 4)/2, (4 + 2)/2) = (3, 3), which lies on y = x. So (4, 2) is the mirror image of (2, 4) in that line, and the same argument works for any point (a, b) and its swap (b, a).

A reflection only looks like one when the two axes have the same scale. On a stretched grid the line y = x is not at 45°, and the two curves stop looking like mirror images.

The graph of $y={\log }_{2}x$y = log₂ x

Reflect every point of $y=2^x$y = 2ˣ in y = x. The points $\left(-2,\frac{1}{4}\right)$(−2, 1/4), $\left(-1,\frac{1}{2}\right)$(−1, 1/2), (0, 1), (1, 2), (2, 4) and (3, 8) become $\left(\frac{1}{4},-2\right)$(1/4, −2), $\left(\frac{1}{2},-1\right)$(1/2, −1), (1, 0), (2, 1), (4, 2) and (8, 3). Each one is a logarithm fact: (8, 3) says ${\log }_{2}8=3$log₂ 8 = 3, and $\left(\frac{1}{2},-1\right)$(1/2, −1) says ${\log }_2\left(\frac{1}{2}\right)=-1$log₂ (1/2) = −1.

Read along the new curve: each time x doubles, y goes up by exactly 1. From x = 1 to 2 it rises 1, and from x = 4 to 8 it also rises only 1, over a much longer stretch. So the curve climbs more and more slowly.

The gold curve is $y={\log }_{2}x$y = log₂ x, the white curve is $y=2^x$y = 2ˣ, and the white straight line is y = x. The pairs (2, 4) and (4, 2), (0, 1) and (1, 0), and $\left(-1,\frac{1}{2}\right)$(−1, 1/2) and $\left(\frac{1}{2},-1\right)$(1/2, −1) are mirror images in y = x.

Every base greater than 1

The same reflection works for any base a greater than 1. Three points on $y={\log }_{a}x$y = logₐ x come straight from the powers of a. Since $a^0=1$a⁰ = 1, the curve passes through (1, 0). Since $a^1=a$a¹ = a, it passes through (a, 1). Since $a^{-1}=\frac{1}{a}$a⁻¹ = 1/a, it passes through $\left(\frac{1}{a},-1\right)$(1/a, −1).

So every logarithm graph crosses the x-axis at x = 1. The base decides how fast it climbs after that: $y={\log }_{10}x$y = log₁₀ x passes through (10, 1) and (0.1, −1), so it needs x = 10 to reach the height that $y={\log }_{2}x$y = log₂ x reaches at x = 2. To the right of x = 1, a bigger base gives a flatter curve: its exponential climbs more steeply, and the reflection of a steeper curve is a flatter one.

The y-axis is an asymptote

No power of 2 is 0, and no power of 2 is negative. So ${\log }_{2}0$log₂ 0 and ${\log }_2\left(-1\right)$log₂ (−1) do not exist, and the graph has no points at all for $x\le 0$x ≤ 0. The domain of $y={\log }_{2}x$y = log₂ x is x > 0.

Close to 0 the curve falls steeply. ${\log }_2\left(\frac{1}{2}\right)=-1$log₂ (1/2) = −1, ${\log }_2\left(\frac{1}{1024}\right)=-10$log₂ (1/1024) = −10, and ${\log }_2$log₂ of one millionth is about −19.93. Each halving of x takes the curve down by 1 more, and x can be halved forever without reaching 0. So the curve runs down alongside the y-axis without ever touching it: the y-axis is a vertical asymptote.

This is the reflection of a fact about $y=2^x$y = 2ˣ. The x-axis is a horizontal asymptote of $y=2^x$y = 2ˣ on the left, and reflecting in y = x turns the x-axis into the y-axis.

$y={\log }_{2}x$y = log₂ x through $\left(\frac{1}{4},-2\right)$(1/4, −2), $\left(\frac{1}{2},-1\right)$(1/2, −1), (1, 0), (2, 1) and (4, 2). To the left of x = 1 each halving of x lowers the curve by 1, and it runs down alongside the y-axis. Nothing is drawn to the left of the y-axis.

No ceiling

To the right the curve keeps climbing, however slowly. ${\log }_{2}1024=10$log₂ 1024 = 10 and ${\log }_{2}1048576=20$log₂ 1048576 = 20, so the curve reaches every height: a height h is reached at x = 2ʰ. It never levels off, so it has no horizontal asymptote. The range of $y={\log }_{2}x$y = log₂ x is every real number, which is the domain of $y=2^x$y = 2ˣ, just as its domain, x > 0, is the range of $y=2^x$y = 2ˣ.

Moving the graph

Transforming Graphs gave the rules: a number added outside the function moves the graph up or down, and a number inside the bracket moves it left or right, the opposite way to its sign. For a logarithm graph, remember to move the asymptote too.

$y={\log }_2\left(x-3\right)$y = log₂ (x − 3) is $y={\log }_{2}x$y = log₂ x moved 3 to the right. The input x − 3 must be positive, so the domain is x > 3, and the asymptote is the line x = 3. The landmarks move 3 right as well: (1, 0) goes to (4, 0), and (2, 1) goes to (5, 1). Check: ${\log }_2\left(4-3\right)={\log }_{2}1=0$log₂ (4 − 3) = log₂ 1 = 0.

$y={\log }_{2}x+2$y = log₂ x + 2 is $y={\log }_{2}x$y = log₂ x moved 2 up. The asymptote is still the y-axis, and the curve crosses the x-axis where ${\log }_{2}x=-2$log₂ x = −2, at $x=\frac{1}{4}$x = 1/4. $y={\log }_2\left(x+1\right)$y = log₂ (x + 1) is $y={\log }_{2}x$y = log₂ x moved 1 to the left, with the asymptote x = −1, and it passes through the origin, since ${\log }_{2}1=0$log₂ 1 = 0.

The white curve is $y={\log }_{2}x$y = log₂ x, through (1, 0) and (2, 1). The gold curve is $y={\log }_2\left(x-3\right)$y = log₂ (x − 3), through (4, 0) and (5, 1): every point has moved 3 to the right, and its asymptote is the dashed line x = 3.

A stretch that is also a translation

A logarithm graph has one more property that other graphs do not. By the product law, ${\log }_2\left(8x\right)={\log }_{2}8+{\log }_{2}x=3+{\log }_{2}x$log₂ (8x) = log₂ 8 + log₂ x = 3 + log₂ x. So squeezing the graph of $y={\log }_{2}x$y = log₂ x horizontally by a factor of 8, which gives $y={\log }_2\left(8x\right)$y = log₂ (8x), draws exactly the same curve as moving it 3 up. Every horizontal stretch of a logarithm graph is a vertical translation.

The usual mistakes

Drawing $y={\log }_{2}x$y = log₂ x through (0, 1). That point belongs to $y=2^x$y = 2ˣ. Swapped, it becomes (1, 0): every logarithm graph crosses the x-axis at x = 1 and never meets the y-axis.

Reading the coordinates the wrong way round. The point (8, 3) on $y={\log }_{2}x$y = log₂ x says ${\log }_{2}8=3$log₂ 8 = 3: the number 8 is across, and its logarithm 3 is up. The logarithm is the power, not the number itself.

Letting the curve cross the y-axis. There is no logarithm of 0 or of a negative number, so nothing is drawn for $x\le 0$x ≤ 0.

Drawing the curve leveling off to a horizontal line. It climbs slowly, but it reaches every height.

Moving $y={\log }_2\left(x-3\right)$y = log₂ (x − 3) to the left. The input is 3 less than x, so the curve reaches each height 3 later, to the right, and its asymptote moves to x = 3.