Introduction to Logarithms Advanced. This lesson goes beyond core Algebra I. You can skip it.

A name for the missing exponent

Look again at $2^x = 10$. The value $2^3 = 8$ falls short and $2^4 = 16$ overshoots, so the exponent we want lies between $3$ and $4$. As $x$ climbs from $3$ to $4$, $2^x$ climbs steadily too, from $8$ up to $16$, without skipping or repeating a value on the way, so it passes through $10$ exactly once. In the exponential functions lesson this steady climb had a name, one-to-one, and it is the reason a definite answer exists even though no whole number gives it. That number has had no symbol until now. We give it one.

Fix a base $b$ with $b > 0$ and $b \ne 1$, and take a positive number $y$. The logarithm of $y$ to base $b$, written $\log_{b}(y)$, is the exponent you put on $b$ to produce $y$. Said as an equation,

The two equations carry the exact same information, read in opposite directions. The right-hand form $b^{x} = y$ is the exponential form; the left-hand form $\log_{b}(y) = x$ is the logarithmic form. One starts from the exponent $x$ and reports the result $y$; the other starts from the result $y$ and reports the exponent $x$. This is the sentence to carry through the whole lesson: a logarithm is an exponent. Whenever you read $\log_{b}(y)$, say to yourself “the exponent that turns $b$ into $y$,” and the symbol stops being mysterious.

So the answer to our opening puzzle finally has a name. The unique solution of $2^x = 10$ is $x = \log_{2}(10)$, a number between $3$ and $4$ that we will pin down to $3.32$ once we have the tools.

Reading the same fact both ways is a skill worth drilling, because every rule that follows is just one of these two forms in disguise:

Reading a logarithm straight from the definition

To evaluate a logarithm by hand, do not reach for a formula. Translate it into the question the definition asks: the base to what power gives the input? For friendly numbers you can answer that question on sight.

The logarithm undoes the exponential

The definition was built to make $\log_{b}$ reverse $b^x$. Apply one after the other, in either order, and you land back where you started: the exponential and the logarithm are a matched pair of undo buttons, the same pattern the inverse functions lesson already named. Written out, that pattern reads

Read each one in words. In $b^{\log_{b}(y)}$, the exponent $\log_{b}(y)$ is by definition “the exponent that turns $b$ into $y$.” So raising $b$ to that exponent turns $b$ into $y$, and you land on $y$. In $\log_{b}(b^{x})$, the input $b^{x}$ is a power of $b$ whose exponent is $x$, and the logarithm reports that exponent, which is $x$.

This inverse relationship also settles a question the definition left hanging: which numbers may you take the logarithm of? A logarithm only ever returns an exponent that the exponential could have used, so its input must be a number the exponential can actually produce. From the exponential functions lesson, $b^{x}$ is always positive: its range is $y > 0$. There is no real power of a positive base that equals $0$ or a negative number. So $\log_{b}(0)$ and the logarithm of any negative number are undefined, and the domain of $\log_{b}$ is the positive numbers alone,

For a base $b > 1$, as the input shrinks toward $0$, the exponent needed to reach it runs off toward $-\infty$. That runaway is why the graph of such a logarithm later dives down along the vertical axis. You can take the logarithm of a tiny positive number, but never of zero, and never of a negative.

Common logs and natural logs

Two bases come up so often that they get their own shorthand.

The common logarithm has base $10$. Because we write numbers in base ten, powers of ten are everywhere, and this log is written with no base at all: $\log(y)$ means $\log_{10}(y)$. So $\log(1000) = 3$ and $\log(100) = 2$. On a calculator it is the button marked $\log$.

The natural logarithm has base $e$, the number $e \approx 2.718$ that surfaced in the compound interest lesson as the ceiling of continuous growth. It is written $\ln(y)$, so $\ln$ means $\log_{e}$. Thus $\ln(e) = 1$ and $\ln(1) = 0$. On a calculator it is the button marked $\ln$. The base $e$ is called “natural” because it is the base that arises on its own in problems of continuous growth and decay. In those problems it makes the formulas simplest.

Everything proved in this lesson holds for every allowed base, including these two. When a statement is written with $\log$ or $\ln$, it is just the general rule with $b = 10$ or $b = e$.

The laws of logarithms

Because a logarithm is an exponent, the rules for combining logarithms are really the rules for combining exponents, wearing a disguise. Each of the three laws below turns a harder operation on the inputs into an easier operation on the logarithms. Look at the pattern on numbers you can verify first, then see why it always holds.

With base $2$, watch what happens when the values multiply. $8 \cdot 4 = 32$, and their logarithms are $\log_{2}(8) = 3$ and $\log_{2}(4) = 2$, which add up to the logarithm of the product: $\log_{2}(8 \cdot 4) = \log_{2}(32) = 5 = 3 + 2$. The same thing happens with powers. Squaring $8$ gives $8^2 = 64 = 2^6$, and $\log_{2}(8^{2}) = 6 = 2 \cdot 3 = 2\log_{2}(8)$. Multiplying the values added their logarithms; raising a value to a power multiplied its logarithm. The figure below shows the doubling pattern in miniature, and the three laws below explain why it always works, not just for $8$ and $4$.

Throughout, fix a base $b$ and two positive numbers $M$ and $N$, and name their logarithms

which in exponential form say $b^{m} = M$ and $b^{n} = N$. Every proof starts from those two equations and applies a single exponent rule you already know.

Step back and notice the pattern across all three laws just proved: multiplication became addition, division became subtraction, and a power became a product. That trade, hard operation for easy one, is not a curiosity; it is the reason logarithms were invented centuries before anyone cared that they were inverse functions. Multiplying two twelve-digit numbers by hand is brutal, but adding their logarithms and looking up the reverse is quick.

Solving for an exponent at last

Now the payoff the whole chapter was building toward. To solve an equation with the unknown stuck in the exponent, such as $2^x = 10$, take the logarithm of both sides. Then let the power law bring that exponent down to the ground, where you can solve for it.

You may take the logarithm in any base you like, as long as it is the same base on both sides. Using base $2$ on $2^x = 10$ gives $x = \log_{2}(10)$ in one step. Most calculators, however, carry only the $\log$ (base $10$) and $\ln$ (base $e$) buttons, so it is handy to use one of those. That same observation gives a way to compute a logarithm in any base from the buttons you have. Solving $b^{x} = y$ by taking the common log of both sides gives $x\log(b) = \log(y)$, so $x = \dfrac{\log(y)}{\log(b)}$. Since $x = \log_{b}(y)$, this is the change-of-base relationship,

which lets you find, say, $\log_{2}(10) = \dfrac{\log(10)}{\log(2)} \approx \dfrac{1}{0.3010} \approx 3.32$ on any calculator.

The graph of a logarithm

Since $y = \log_{b}(x)$ is the inverse of $y = b^{x}$, its graph is not something new to plot point by point. By the graphs of inverse functions lesson, the graph of an inverse is the mirror image of the original across the line $y = x$. Reflect the exponential curve across that diagonal and you have the logarithm.

The reflection swaps the coordinates of every point, so it turns each feature of the exponential into its mirror feature. Line them up:

• The exponential passes through $(0, 1)$, so the logarithm passes through $(1, 0)$. Indeed $\log_{b}(1) = 0$. The exponential also passes through $(1, b)$, so the logarithm passes through $(b, 1)$, matching $\log_{b}(b) = 1$.
• The exponential has the horizontal asymptote $y = 0$; reflected, the logarithm has the vertical asymptote $x = 0$. The curve dives down the $y$-axis but never touches it.
• The exponential has domain all real numbers and range $y > 0$. Swapping the two, the logarithm has domain $x > 0$ and range all real numbers. You can only take the log of a positive number, but the output can be any real number at all.
• The exponential with base $b > 1$ is increasing, and its reflection is increasing too, so $y = \log_{b}(x)$ rises from left to right, slowly but without bound.

Three settings of that base are worth visiting deliberately. Base $3$ draws the curve the next worked example describes, so you can check your answer against the picture. Base $\tfrac{1}{2}$ is the case the bullets above do not cover. A base below $1$ makes the exponential fall, and reflecting a falling curve gives a falling curve, so the logarithm decreases instead of increasing. Everything else survives unchanged, because the reflection does not care which way the original ran. Even with a base below $1$, the logarithm still passes through $(1, 0)$, still has the vertical asymptote $x = 0$, and still accepts only positive inputs.

Base $1$ is the setting where the logarithm vanishes from the figure entirely, and that is the definition being enforced rather than a fault in the drawing. Every power of $1$ is $1$, so $y = 1^x$ is the flat line $y = 1$. Reflect that line across $y = x$ and you get the vertical line $x = 1$, but a vertical line fails the vertical-line test: it assigns more than one output to the same input, so it is not the graph of a function at all. Since $y = 1^x$ sends every input to the same output, no rule can undo it, and its mirror image cannot be a logarithm. That is precisely what the condition $b \ne 1$ in the definition was written to exclude.

Common mistakes