Treat the log as a function, y = loga x — the mirror image of y = ax across the line y = x.
A logarithm returns an exponent. In the last lesson, Logarithms, we read ax = N backward to get x = loga N — the power that builds N. Now feed that machine every positive input and it becomes a function, y = loga x. Its graph is not something new to memorize: it is the exponential y = ax seen in a mirror, and the mirror is the line y = x. This lesson draws the log curve, lists what it always does — lives only for x > 0, passes through (1, 0), climbs ever more slowly — and shows exactly why exponential and logarithm reflect into each other and cancel one another out.
Fix a base — take a = 2 first, the friendly doubling base. The rule y = log2 x reads: "y is the power of 2 that gives x." To plot it, we do not need a single new idea — we just turn the exponential table on its side. We already know that 2 to the power 0 is 1, to the power 1 is 2, to the power 2 is 4, and so on; so the logarithm reverses each line:
| x | ¼ | ½ | 1 | 2 | 4 | 8 |
|---|---|---|---|---|---|---|
| y = log2 x | −2 | −1 | 0 | 1 | 2 | 3 |
Read the row left to right. As x shrinks toward 0 — ¼, ⅛, ¹⁄₁₆ — the output dives to −2, −3, −4 and keeps falling without bound: the curve plunges down along the y-axis but never touches it. As x grows, the output climbs, but ever more slowly: it takes a doubling of x to add just 1 to y. Joining the dots gives a smooth curve that crosses the x-axis at (1, 0) (because 20 = 1) and passes through (2, 1) (because 21 = 2). In general it passes through (a, 1) for any base a, since loga a = 1.
The most important feature is what the curve refuses to do. A logarithm only eats positive numbers: there is no power of 2 that gives 0 or a negative number, so log2 x is undefined for x ≤ 0. The whole graph lives strictly to the right of the y-axis, and the line x = 0 is a vertical asymptote — drawn in red below, a wall the curve presses against forever but never reaches.
y = loga x (with a > 0, a ≠ 1) is defined only for x > 0. It always passes through (1, 0) and (a, 1), and the line x = 0 (the y-axis) is a vertical asymptote it never crosses.
Move the base and the curve breathes but keeps its shape. Try it: dial a through 2, 3, 5 (steep-then-flatter rising curves), and through ½ and ⅓ (falling curves). Every one of them threads the same anchor (1, 0) and clings to the same red wall.
Strip away the particular base and four facts hold for every y = loga x.
Domain. The inputs are exactly the positive numbers, x > 0 — in interval language (met in Intervals and neighborhoods), the open ray (0, ∞). Nothing to the left of, or on, the y-axis is allowed.
Range. The outputs sweep through all real numbers, (−∞, ∞): push x toward 0 and y dives without bound; push x large and y climbs without bound. (This is the exact opposite of the exponential, whose range was y > 0 — the two simply trade their domain and range, as a mirror should.)
The anchor. Every log curve passes through (1, 0), because a0 = 1 for every base, so loga 1 = 0.
Direction. When a > 1 the curve increases (rises left to right); when 0 < a < 1 it decreases (falls). And in both cases the y-axis, x = 0, is a vertical asymptote.
| a > 1 (e.g. log2 x) | 0 < a < 1 (e.g. log½ x) | |
|---|---|---|
| domain | x > 0, i.e. (0, ∞) | x > 0, i.e. (0, ∞) |
| range | all reals (−∞, ∞) | all reals (−∞, ∞) |
| passes through | (1, 0) and (a, 1) | (1, 0) and (a, 1) |
| behaviour | increasing ↗ | decreasing ↘ |
| asymptote | x = 0 (vertical) | x = 0 (vertical) |
The wall here is vertical: x = 0, not the horizontal y = 0 of the exponential. A common slip is to "remember the asymptote" from Exponential Functions and place it along the x-axis. For a log, the curve does cross the x-axis (at (1, 0)); it is the y-axis it never touches.
Why does the log curve look like the exponential turned on its side? Because the two functions ask opposite questions about the same fact ax = N. The exponential y = ax takes the exponent and returns the result; the logarithm y = loga x takes the result and returns the exponent. Each function undoes the other — they are inverse functions.
Inverse functions have a beautiful graphical signature. If raising 2 to the power 3 gives 8, then the point (3, 8) sits on y = 2x; and "the power of 2 that gives 8 is 3" puts the point (8, 3) on y = log2 x. The two points are the same pair of numbers with their coordinates swapped. And swapping the x- and y-coordinates of a point is exactly what reflecting it across the line y = x does. So:
(p, q) lies on y = ax ⟺ (q, p) lies on y = loga x.
Fold the whole exponential curve across y = x and it lands exactly on the logarithm. The figure below proves it honestly: the green log curve is drawn as the reflection of the blue exponential — same points, x and y swapped — and the dashed connectors between matching points all cross y = x at right angles, which is the geometric fingerprint of a mirror. The special points fold as you would expect: (0, 1) ↔ (1, 0) and (1, a) ↔ (a, 1).
To reflect a point across y = x, swap its coordinates: (p, q) → (q, p). Because y = ax and y = loga x are inverse functions, their graphs are mirror images across y = x.
If two functions undo each other, then doing one and then the other returns you to where you started — like flipping a switch off, then on. In symbols, for the exponential and the logarithm with the same base a:
aloga x = x (for x > 0) and loga(ax) = x (for all x).
Read the first one slowly. The inside, loga x, is "the power of a that gives x." Raise a to that power and of course you get x back — that is what the exponent was chosen to do. The second is the same idea from the other side: raise a to the x, then ask which power of a that was; the answer is x. Logarithm and exponential are a perfectly matched lock and key.
• 2log2 5 = 5 — the exponent log2 5 is whatever power of 2 makes 5,
so 2 raised to it is 5.
• log2(27) = 7 — log2 of 2-to-a-power just reads the power off.
• 10lg 3 = 3 and log5(54) = 4 — same cancellation, any base.
The widget below makes the switch literal: pick any x, and watch both round trips land you right back on x.
The cancellation only means something where each side is defined. aloga x = x needs x > 0 (you cannot take log of a non-positive number to begin with), while loga(ax) = x works for all x, because ax is always positive. Never write something like "log2(2x) = x" and then plug in a negative x to the wrong identity — keep track of which switch you are flipping.
Because a log curve with a > 1 is increasing, it answers "which is bigger?" by position: the input farther to the right sits higher. So
log2 5 < log2 7 (since 5 < 7, and log2 x increases).
Reading y = log2 x against the x-axis splits the inputs into three honest regions. For a base greater than 1:
| region of x | 0 < x < 1 | x = 1 | x > 1 |
|---|---|---|---|
| sign of loga x (a > 1) | negative | zero | positive |
Why? A number between 0 and 1, like ½, is a negative power of a (½ = 2−1), so its log is negative; at x = 1 the power is 0; beyond 1 the power is positive. For a base with 0 < a < 1 the curve is decreasing, so every one of these signs flips: there, larger inputs give smaller outputs.
Use the base slider from §24.4.1 again — it reports loga of two sample inputs so you can watch the order. With a = 2 it shows log2 5 ≈ 2.32 below log2 7 ≈ 2.81; switch to a = ½ and the same two inputs come back in the opposite order, because the curve now falls.
For a > 1: loga x is negative on 0 < x < 1, zero at x = 1, positive for x > 1, and larger inputs give larger outputs. For 0 < a < 1 every comparison reverses.
The logarithmic function is the exponential in a mirror. Treat the logarithm as a function, y = loga x (a > 0, a ≠ 1): it is defined only for x > 0, its range is all real numbers, it always passes through (1, 0) and (a, 1), and the y-axis x = 0 is a vertical asymptote it never crosses. It increases when a > 1 and decreases when 0 < a < 1. Because y = ax and y = loga x are inverse functions, their graphs are reflections across y = x — every point (p, q) on one becomes (q, p) on the other — and they cancel: aloga x = x and loga(ax) = x. Next, in Growth Models, we aim all of this at the world: compound interest, half-life, and using a log to free an exponent.
What is the domain of y = log3 x? Write it as an interval.
A logarithm only accepts positive inputs, so the domain is x > 0, that is the interval (0, ∞). (The y-axis x = 0 is a vertical asymptote.)
Through what point does every graph y = loga x pass, no matter the base?
(1, 0) — because a0 = 1 for every base, so loga 1 = 0.
The point (3, 1) lies on the curve y = ax (so a3 = 1 here would be wrong — read it as: at x = 3, y = 1 is just an example point). Reflect (3, 1) across y = x: what point lands on the log curve?
Reflecting across y = x swaps the coordinates: (3, 1) → (1, 3). So (1, 3) lies on the matching logarithmic curve y = loga x.
Simplify log5(54).
Logarithm and exponential of the same base cancel: log5(54) = 4 — log5 just reads the exponent off.
Simplify 7log7 2.
The exponent log7 2 is "the power of 7 that gives 2," so 7 raised to it is exactly 2. (This needs 2 > 0, which it is.)
Which is larger, log2 10 or log2 20? Explain in one line.
log2 20 is larger. Since the base 2 > 1, the function log2 x is increasing, so a bigger input gives a bigger output (20 > 10 ⇒ log2 20 > log2 10). Numerically, log2 10 ≈ 3.32 and log2 20 ≈ 4.32.
Six questions to lock it in. Tap the answer you think is right.
The big idea. A logarithmic function is not a new object to memorize — it is the exponential function read backward, and its graph is the reflection of y = ax across the line y = x. If a student firmly grasps "swap the coordinates," the domain (x > 0), the anchor (1, 0), the asymptote (x = 0), and the increasing/decreasing rule all follow from facts they already know about exponentials in 24.2. The cancellation identities aloga x = x and loga(ax) = x are the algebraic face of that same inverse relationship.
Misconceptions to watch. (1) Thinking loga x is defined for all x — it lives only for x > 0. (2) Confusing the asymptote: for a log it is the vertical line x = 0, not the horizontal y = 0 of the exponential; the log does cross the x-axis, at (1, 0). (3) Forgetting that reflection across y = x swaps coordinates, so a point (3, 8) on the exponential becomes (8, 3) — not (3, 8) — on the log.
Common Core. This lesson supports F-BF.B.4 (find inverse functions and relate their graphs), F-IF.C.7e (graph logarithmic functions, showing intercepts and asymptotic behavior), and F-LE.A.4 (use logarithms to express the solution of an exponential equation; relate logs and exponents).