g(x) = log_b(x) is the inverse of f(x) = b^x, so its graph is that curve reflected in the line y = x. The domain, the range, the asymptote and the slowness of the climb are all consequences of the reflection rather than separate facts to memorise.
Core ideas
One picture and one table serve both functions. Read the table down for the exponential and across for the logarithm.
If (p, q) lies on the graph of an invertible function, (q, p) lies on the graph of its inverse. 2³ = 8 puts (3, 8) on y = 2^x and (8, 3) on y = log_2(x). There is one set of pairs here, read in two directions.
b^x accepts every real number and returns only positive ones. So log_b(x) accepts only positive numbers and returns every real one. The exclusion of x ≤ 0 is not an extra rule bolted on; it is the range of the exponential seen from the other side.
y = b^x flattens toward the horizontal line y = 0. Reflecting that line in y = x gives the vertical line x = 0, and that is the logarithm's asymptote: the curve plunges without bound as x comes down toward 0 from the right.
The material
Why the reflection is the whole picture, and what can be read off it before anything is plotted.
Lines, Functions and Graphs defined an inverse function: g undoes f when g(f(x)) = x and f(g(x)) = x, and a function has one only if it is one-to-one. f(x) = b^x is strictly increasing when b > 1 and strictly decreasing when 0 < b < 1, so it never repeats a value and the inverse exists without any restriction of the domain — which is unusual, and is why logarithms are as well-behaved as they are.
For b > 0 with b ≠ 1, the logarithmic function with base b is g(x) = log_b(x), defined for x > 0. It is the inverse of f(x) = b^x: log_b(b^x) = x for every real x, and b^(log_b(x)) = x for every x > 0.
Both halves of that sentence matter. To lift log_10(x) by one you must multiply x by ten, so the output crawls while the input explodes. But there is no height it fails to reach eventually.
That last line is the reason decibels, pH and earthquake magnitude are logarithms, which is “Logarithmic Scales”. It is also why a logarithmic curve looks flat on any drawing wide enough to be interesting — flat is not level, and the two are easy to confuse by eye.
log_b(x − h) is the same curve moved h to the right, and the asymptote goes with it to x = h. The domain becomes x > h. This is the commonest source of a wrong domain in the whole course, because the shift is invisible unless you look inside the brackets.
The argument must be positive: x − 5 > 0, so the domain is x > 5 and the asymptote is the vertical line x = 5. The curve passes through (6, 0), because x − 5 = 1 there and log_3(1) = 0, and through (8, 1), because x − 5 = 3 and log_3(3) = 1. It is the standard shape, five units to the right.
One case worth having met early: the domain is decided by the argument being positive, not by x being positive. log_2(x²) is defined for every x ≠ 0, negative values included, because x² is positive there. “The Laws of Logarithms” shows why that makes one of the three laws need a careful statement.
Interactive
y = b^x and y = log_b(x) across the line y = x
Method
Four marks on the page, in this order, and the shape is fixed.
x = 0, or x = h if the argument is x − h. Putting it down before the curve stops the sketch from wandering into a region where the function does not exist.
(1, 0) and (b, 1) lie on every logarithmic curve, straight from the definition. For base 3 they are (1, 0) and (3, 1); shift both if the argument is shifted.
(1/b, −1). Without it the sketch tends to stop at the axis instead of diving, and the diving part is the half that carries the asymptote.
x > 0, or whatever makes the argument positive. A logarithmic graph without its domain is the half of the answer that gets marked wrong, and it is the half “Solving Logarithmic Equations” depends on.
The first block is one table read twice. The second is what happens to every one of those facts when the argument changes.
The domain x > 5 is not a decoration on the answer. log_2(2 − 5) = log_2(−3) names no real number, so the curve has nothing at all to the left of 5, and a candidate solution of x = 2 to any equation built from this expression is not a solution. “Solving Logarithmic Equations” is built on precisely that check.
Note that the shift moved the two free points too. (1, 0) became (6, 0) and (2, 1) became (7, 1). Nothing about the shape changed; every landmark moved five to the right, the asymptote included.
For a faded sketch, use g(x) = log_4(x + 3). The first line is supplied: x + 3 > 0. Finish the domain, draw the vertical asymptote, move the three anchor inputs 1, 4 and 1/4 through the shift, and state the direction. Check each plotted point by converting it back to an exponential equation.
Given y = log_5(x + 2) you should say: domain x > −2, asymptote x = −2, through (−1, 0) and (3, 1), increasing. If you need a table of values to get there, the reflection has not landed as a fact you can use.
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