Introduction to Logarithms

Why the exponential can be run backward

Fix a base $b$ with $b > 0$ and $b \ne 1$, and consider the exponential function $f(x) = b^x$. From the previous lesson you know three things about it. It is defined for every real number $x$. Its outputs are always positive. And its graph is an unbroken curve that rises steadily when $b > 1$ and falls steadily when $0 < b < 1$.

“Rises steadily” is the part that matters, because it is what makes the function one-to-one. It deserves a proof rather than a glance at a picture, since everything in this lesson rests on it.

Being one-to-one tells you the inverse exists. To know what to feed it, you need the range of $b^x$, because the Inverse Functions lesson pinned the two down together. The inverse’s domain is the original function’s range, and the inverse’s range is the original function’s domain.

Every power of a positive base is positive, so $b^x$ never outputs $0$ and never outputs a negative number. Those values are simply not available. On the other side, with $b > 1$ the curve climbs past every bound as $x$ grows. Still with $b > 1$, as $x$ runs off to the left, $b^x = \tfrac{1}{b^{-x}}$ shrinks toward $0$ without ever landing on it. An unbroken increasing curve that gets arbitrarily close to $0$ at one end and arbitrarily large at the other passes through every height in between. Such a curve reaches each of those heights exactly once, because it never repeats a value.

A base with $0 < b < 1$ needs no fresh argument, just the reduction the proof already used. Write $b = \tfrac{1}{c}$ with $c > 1$, so that $b^x = c^{-x}$. As $x$ runs over every real number, so does $-x$, so $b^x$ produces exactly the same collection of outputs that $c^x$ does, merely in the reverse order. Its range is the positive numbers too. Either way, then:

Read that through the inverse rule and the inverse is already fully described before it has a name. Its inputs are the positive numbers, its outputs are all the real numbers, and the number it returns for a given input is an exponent. All that is left is to write it down.

The logarithm is the name of that inverse

Here is the definition, and it is the whole lesson in one sentence.

Fix a base $b$ with $b > 0$ and $b \ne 1$. For a positive number $x$, the logarithm of $x$ to the base $b$, written $\log_b x$, is the unique real exponent you must put on $b$ to get $x$.

Unpacked into an equation, the definition is a two-way street between a logarithmic statement and an exponential one:

Both directions are true, and both are worth checking. Going left to right, if $y = \log_b x$ then $y$ is by definition the exponent that turns $b$ into $x$, so $b^y = x$. Going right to left, suppose $b^y = x$ for some real $y$. Then $x$ is automatically positive, so it is a legal input for the logarithm, and $y$ is an exponent that turns $b$ into $x$. There cannot be a second such exponent, because $b^t$ is one-to-one in $t$, so $y$ has to be the one the logarithm names, and $y = \log_b x$. The equivalence goes both ways, which is why you may swap either form for the other at any moment without losing information.

Say the definition aloud whenever you use it. "$\log_2 8$" means ”$2$ to the what gives $8$?” The answer is $3$, and so $\log_2 8 = 3$.

The two restrictions on the base are not decoration. If $b = 1$, then $1^y = 1$ for every exponent $y$, so the function is a flat line: it is not one-to-one, and it never produces any output except $1$. The equation $1^y = 5$ has no solution and $1^y = 1$ has infinitely many, so no inverse can exist. If $b$ were negative, $b^x$ would not even be a function on the real numbers, since a power like $(-4)^{1/2}$ is not a real number at all. There is nothing there to invert.

Two values come free from the definition and are worth memorising on the spot:

Reading a logarithm straight from the definition

Every evaluation is the same question asked with different numbers, so ask it out loud each time.

When the base and the number are powers of a common smaller number, write both as powers of that number and the exponent falls out.

The two cancellation laws

The Inverse Functions lesson ended with a pair of identities: composing a function with its inverse, in either order, gives back what you started with. Written for $f(x) = b^x$ and $f^{-1}(x) = \log_b x$, they read

Neither needs the general theory to justify it; each is one line from the definition. For the first, ask what $\log_b(b^t)$ means: it is the exponent you put on $b$ to get $b^t$. The exponent $t$ does that job, and one-to-one-ness says no other exponent does, so the answer is $t$. For the second, $\log_b x$ is by construction the exponent that turns $b$ into $x$; put it on $b$, and you get $x$.

Watch the two domains, because they are not the same. You may write $\log_b(b^t) = t$ for every real $t$, positive or negative, since $b^t$ is a legal input for the logarithm no matter what $t$ is. But $b^{\log_b x} = x$ requires $x > 0$, because otherwise $\log_b x$ does not exist and the left side is meaningless. Written as $\log_7\!\left(7^{-5}\right) = -5$ and $7^{\log_7 12} = 12$, the two laws are easy to use and easy to confuse, so keep asking which one you are applying.

The cancellation laws combine with the exponent rules to settle expressions that look far worse than they are. Take $9^{\log_3 5}$. The base $9$ is $3^2$, so

Nothing there is a new rule. It is $\left(b^m\right)^n = \left(b^n\right)^m$ followed by the cancellation law, and the tangle collapses to $25$.

What you are allowed to take the logarithm of

The domain question was answered before the logarithm even had a name. Its domain is the range of $b^x$, and its range is the domain of $b^x$:

So $\log_2 0$ and $\log_2(-8)$ are not small, or negative, or infinite. They do not exist. There is no real exponent $y$ with $2^y = 0$, because powers of $2$ stay strictly positive however far left you go, and there is certainly no exponent with $2^y = -8$. For a fixed legal base, $\log_b x$ is defined exactly when $x > 0$: positive inputs always work, since every positive number is $b$ to some power, and nothing else ever works.

In practice this restriction bites on the argument, the quantity sitting inside the logarithm, not on the variable. The function $y = \log_5(x - 2)$ is defined precisely when $x - 2 > 0$, which is to say $x > 2$. The function $y = \log_2(x^2 + 1)$ is defined for every real $x$, negative $x$ included, because $x^2 + 1$ is positive no matter what. Always ask what the logarithm is being handed, and then demand that that be positive.

The graph is the exponential reflected across the line $y = x$

Inverse functions swap coordinates. The point $(a, c)$ lies on the graph of $f$ exactly when $(c, a)$ lies on the graph of $f^{-1}$. That coordinate swap is the same thing as reflecting the plane across the line $y = x$. So you do not need a new table of values to graph a logarithm. You need the exponential’s table, read backward.

Every feature of the logarithm’s graph is now something you can read off the reflection rather than memorise.

The exponential’s $y$-intercept $(0, 1)$ becomes the logarithm’s $x$-intercept $(1, 0)$. So the graph of $y = \log_b x$ crosses the $x$-axis at $x = 1$ for every base, which is just $\log_b 1 = 0$ drawn as a picture. The point $(1, b)$ on the exponential becomes $(b, 1)$ on the logarithm, which is $\log_b b = 1$.

That last pair is the one marked in the figure above, and it is worth stepping the base through every setting to watch it travel. The two marks slide in opposite directions along their curves, always the same distance from the diagonal on opposite sides. Set the base to $1$ and the logarithm disappears altogether. That is not a gap in the drawing. Every power of $1$ is $1$, so $y = 1^x$ is the flat line $y = 1$, one height reached by every input. A rule that sends everything to one place has nothing to undo it. The condition $b \ne 1$ in the definition is exactly this state being excluded.

The exponential’s horizontal asymptote $y = 0$ reflects into a vertical asymptote $x = 0$ for the logarithm. The curve dives down the side of the $y$-axis, getting arbitrarily low without ever touching the axis. That is the picture of “you cannot take the logarithm of $0$”: there is no height at all above $x = 0$.

Reflecting an increasing curve gives an increasing curve, so $\log_b x$ is increasing when $b > 1$ and decreasing when $0 < b < 1$, matching its exponential exactly. And the growth is slow. If $\log_2 x = t$ then $x = 2^t$, so doubling the input gives $2x = 2^{t+1}$, an input whose logarithm is only $t + 1$. Each doubling of $x$ buys a single unit of height. That is why $\log_2 1000$ is under $10$, and why the right-hand end of the curve looks almost flat.

Switching forms to solve equations

The equivalence $y = \log_b x \iff b^y = x$ is a tool, not just a slogan. Whenever an unknown is trapped in an awkward position, translate the statement into the other form and the unknown steps into the open. There are only three places it can hide.

If the unknown is inside the logarithm, as in $\log_2 x = 5$, go to exponential form: $x = 2^5 = 32$.

If the unknown is the base, as in $\log_x 81 = 4$, go to exponential form again: $x^4 = 81$. Now $x = 3$ and $x = -3$ both satisfy that equation, but a logarithm base must be positive and cannot be $1$, so $x = 3$ is the only answer. The rejected root is not a mistake in the algebra; it is a value the definition forbids.

If the unknown is the exponent, as in $3^x = \tfrac{1}{81}$, go to logarithmic form: $x = \log_3 \tfrac{1}{81} = -4$. This is the case the logarithm was built for, and it is why the whole lesson matters.

Trapping a logarithm between two integers

Most logarithms are not tidy. No whole number answers ”$2$ to the what gives $10$?”, and no fraction does either. Suppose one did, say $\log_2 10 = \tfrac{p}{q}$ for positive integers $p$ and $q$. Then $2^{p/q} = 10$, and raising both sides to the $q$th power gives $2^p = 10^q = 2^q \cdot 5^q$. Dividing through by $2^q$ leaves $5^q = 2^{p-q}$. The left side is an odd integer bigger than $1$. The right side is $2$ to an integer power, so it is a fraction below $1$ when $p < q$, exactly $1$ when $p = q$, and an even integer when $p > q$. It is never an odd integer bigger than $1$, so no such $p$ and $q$ exist, and $\log_2 10$ is irrational.

That does not make it mysterious. You can pin it between two integers using nothing but powers of $2$ and the fact that the logarithm is increasing.

First, why is $\log_2$ increasing? Suppose $0 < u < v$ but $\log_2 u \ge \log_2 v$. Raising $2$ to both sides of that inequality preserves it, since $2^x$ is increasing, and the cancellation law turns each side back into the original number, giving $u \ge v$. That contradicts $u < v$. So $u < v$ forces $\log_2 u < \log_2 v$, and applying a base-$2$ logarithm to an inequality keeps it pointing the same way.

Now trap the number. Since $2^3 = 8$ and $2^4 = 16$,

So $\log_2 10$ sits between $3$ and $4$, and its true value is about $3.32$. Do not try to guess where in that interval it lands by asking whether $10$ is nearer to $8$ or to $16$. Arithmetic nearness is the wrong measure, because the height climbs by one unit every time the argument doubles, not every time it grows by $8$. The exponent reaches the halfway height $3.5$ at

so the midpoint of the interval sits above $10$, not at the arithmetic middle $12$. Since $10 < 11.3$ and the logarithm is increasing, $\log_2 10$ lies in the lower half of the interval. Even without that refinement, knowing which two integers a logarithm lives between is enough to sanity-check almost any answer.

Base $10$ deserves a note of its own. It is used so often that it is written with no base at all, and $\log x$ with a bare “log” means $\log_{10} x$ from here on. The trapping argument is especially sharp there. A whole number $n$ with $d$ digits satisfies $10^{d-1} \le n < 10^{d}$, so applying $\log$ gives

In other words, the whole-number part of a common logarithm is one less than the digit count. Since $1000 \le 4500 < 10000$, you know without any calculation that $\log 4500$ is three point something, and sure enough $4500$ has four digits.

What you cannot do yet is compute $\log_2 10$ or $\log 4500$ to a decimal, or break a complicated logarithm into simpler ones. That takes rules for combining logarithms, which is the business of the next lesson. Everything in this one came from a single idea: the logarithm is the exponential run backward.

Common mistakes