What Is a Logarithm?

What Is a Logarithm? #

Our journey begins in the late 1500s and early 1600s. Mathematicians of the era faced a key challenge, one with which I'm sure all of us are familiar: multiplication is harder than addition. What do I mean by harder? Well, consider adding two multiple-digit numbers: $243$ and $142\text{.}$ The standard algorithm is to line these numbers up and add their pairs of digits individually from right to left.

\begin{array}{r} 243 \\ +\;142 \\ \hline \phantom{385} \end{array} \qquad \begin{array}{r} 24{\color{58c4dd} 3} \\ +\;14{\color{58c4dd} 2} \\ \hline \phantom{38}{\color{58c4dd} 5} \end{array} \qquad \begin{array}{r} 2{\color{83c167} 4}{\color{58c4dd} 3} \\ +\;1{\color{83c167} 4}{\color{58c4dd} 2} \\ \hline \phantom{3}{\color{83c167} 8}{\color{58c4dd} 5} \end{array} \qquad \begin{array}{r} {\color{ffee00} 2}{\color{83c167} 4}{\color{58c4dd} 3} \\ +\;{\color{ffee00} 1}{\color{83c167} 4}{\color{58c4dd} 2} \\ \hline {\color{ffee00} 3}{\color{83c167} 8}{\color{58c4dd} 5} \end{array}

\begin{array}{r} 243 \\ +\;142 \\ \hline \phantom{385} \end{array} \qquad \begin{array}{r} 24{\color{0000cc} 3} \\ +\;14{\color{0000cc} 2} \\ \hline \phantom{38}{\color{0000cc} 5} \end{array} \qquad \begin{array}{r} 2{\color{008000} 4}{\color{0000cc} 3} \\ +\;1{\color{008000} 4}{\color{0000cc} 2} \\ \hline \phantom{3}{\color{008000} 8}{\color{0000cc} 5} \end{array} \qquad \begin{array}{r} {\color{ff5500} 2}{\color{008000} 4}{\color{0000cc} 3} \\ +\;{\color{ff5500} 1}{\color{008000} 4}{\color{0000cc} 2} \\ \hline {\color{ff5500} 3}{\color{008000} 8}{\color{0000cc} 5} \end{array}

Additions can get a bit more difficult if we need to carry a $1\text{,}$ but overall this process is pretty straightforward.

Multiplication, on the other hand, involves more steps. Let's take $243$ and $142$ once again and multiply them.

\begin{array}{r} 243 \\ \cdot\;142 \\ \hline \phantom{M} \\ \phantom{M} \\ \phantom{M} \\ \phantom{M} \end{array} \qquad \begin{array}{r} {\color{58c4dd} 243} \\ \cdot\;14{\color{58c4dd} 2} \\ \hline {\color{58c4dd} 486} \\ \phantom{M} \\ \phantom{M} \\ \phantom{M} \end{array} \qquad \begin{array}{r} {\color{83c167} 243} \\ \cdot\;1{\color{83c167} 4}{\color{58c4dd} 2} \\ \hline {\color{58c4dd} 486} \\ {\color{83c167} 972}0 \\ \phantom{M} \\ \phantom{M} \end{array} \qquad \begin{array}{r} {\color{ffee00} 243} \\ \cdot\;{\color{ffee00} 1}{\color{83c167} 4}{\color{58c4dd} 2} \\ \hline {\color{58c4dd} 486} \\ {\color{83c167} 972}0 \\ {\color{ffee00} 243}00 \\ \phantom{M} \end{array} \qquad \begin{array}{r} 243 \\ \cdot\;142 \\ \hline {\color{fc6255} 486} \\ {\color{fc6255} 9720} \\ +\;{\color{fc6255} 24300} \\ \hline {\color{fc6255} 34506} \end{array}

\begin{array}{r} 243 \\ \cdot\;142 \\ \hline \phantom{M} \\ \phantom{M} \\ \phantom{M} \\ \phantom{M} \end{array} \qquad \begin{array}{r} {\color{0000cc} 243} \\ \cdot\;14{\color{0000cc} 2} \\ \hline {\color{0000cc} 486} \\ \phantom{M} \\ \phantom{M} \\ \phantom{M} \end{array} \qquad \begin{array}{r} {\color{008000} 243} \\ \cdot\;1{\color{008000} 4}{\color{0000cc} 2} \\ \hline {\color{0000cc} 486} \\ {\color{008000} 972}0 \\ \phantom{M} \\ \phantom{M} \end{array} \qquad \begin{array}{r} {\color{ff5500} 243} \\ \cdot\;{\color{ff5500} 1}{\color{008000} 4}{\color{0000cc} 2} \\ \hline {\color{0000cc} 486} \\ {\color{008000} 972}0 \\ {\color{ff5500} 243}00 \\ \phantom{M} \end{array} \qquad \begin{array}{r} 243 \\ \cdot\;142 \\ \hline {\color{d94234} 486} \\ {\color{d94234} 9720} \\ +\;{\color{d94234} 24300} \\ \hline {\color{d94234} 34506} \end{array}

Again we start by lining up the numbers. But now we need to multiply the top number by every digit in the bottom number. And we have to shift the results as we go along, then add everything together at the end.

Phew! That was a lot of work. It's nothing new or difficult, just rather tedious. But more than that, we have lots of opportunities for making mistakes. Keeping track of all of the digits and carries and shifts by hand requires a good deal of discipline and attention to detail. Wouldn't it be great if we didn't have to do all of this? Well, that's where logarithms are going to help us!

Logarithms first were created as a tool to avoid multiplications. As the centuries have gone on we've learned that logarithms have a myriad of other mathematical properties. But this is where we'll start: simple addition and multiplication. To begin, we'll reexamine some of the arithmetic we learned in elementary school—doing so will help us motivate the definition of a logarithm.

It All Starts With Counting #

Let's think back to our early educations. First we learned about counting, then about addition and subtraction, and then about multiplication, where we were introducted to evaluating expressions such as ${{\color{58c4dd} 3} \cdot {\color{83c167} 5}}\text{.}$ ${{\color{0000cc} 3} \cdot {\color{008000} 5}}\text{.}$ Now if your first instinct is to yell out ${\color{ffee00} 15}\text{,}$ ${\color{ff5500} 15}\text{,}$ then congratulations—you memorized your times tables when you were a child! If you're anything like me, then you had to learn the products of single-digit number pairs.

\begin{array}{c|rrrrrrrrrr} \cdot & \phantom{8}0 & \phantom{8}1 & \phantom{8}2 & \phantom{8}3 & \phantom{8}4 & \phantom{8}5 & \phantom{8}6 & \phantom{8}7 & \phantom{8}8 & \phantom{8}9 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 0 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 \\ 3 & 0 & 2 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\ 4 & 0 & 2 & 8 & 12 & 16 & 20 & 24 & 28 & 32 & 36 \\ 5 & 0 & 2 & 10 & 15 & 20 & 25 & 30 & 25 & 40 & 45 \\ 6 & 0 & 2 & 12 & 18 & 24 & 30 & 36 & 42 & 48 & 54 \\ 7 & 0 & 2 & 14 & 21 & 28 & 35 & 42 & 49 & 56 & 63 \\ 8 & 0 & 2 & 16 & 24 & 32 & 40 & 48 & 56 & 64 & 72 \\ 9 & 0 & 2 & 18 & 27 & 36 & 45 & 54 & 63 & 72 & 89 \\ \end{array}

But what if we didn't have this table? How would we know what to do when faced with a multiplication problem? Well, we probalby learned that multiplication is a repeated addition. If we're given ${{\color{58c4dd} 3} \cdot {\color{83c167} 5}}\text{,}$ ${{\color{0000cc} 3} \cdot {\color{008000} 5}}\text{,}$ we can write

{\color{58c4dd} 3} \cdot {\color{83c167} 5} = \underbrace{{\color{58c4dd} 3} + {\color{58c4dd} 3} + {\color{58c4dd} 3} + {\color{58c4dd} 3} + {\color{58c4dd} 3}}_{{\color{83c167} 5}\text{ times}} = {\color{ffee00} 15}.

{\color{0000cc} 3} \cdot {\color{008000} 5} = \underbrace{{\color{0000cc} 3} + {\color{0000cc} 3} + {\color{0000cc} 3} + {\color{0000cc} 3} + {\color{0000cc} 3}}_{{\color{008000} 5}\text{ times}} = {\color{ff5500} 15}.

We add ${\color{58c4dd} 3}$ ${\color{0000cc} 3}$ to itself ${\color{83c167} 5}$ ${\color{008000} 5}$ times, and our answer is the total sum of ${\color{ffee00} 15}\text{.}$ ${\color{ff5500} 15}\text{.}$ The multiplication consists of three components: a ${\color{58c4dd} \text{number}}\text{,}$ ${\color{0000cc} \text{number}}\text{,}$ a ${\color{83c167} \text{count}}\text{,}$ ${\color{008000} \text{count}}\text{,}$ and a ${\color{ffee00} \text{total}}\text{.}$ ${\color{ff5500} \text{total}}\text{.}$ To be a bit more general, we can write

{\color{58c4dd} \text{number}} \cdot {\color{83c167} \text{count}} = \underbrace{{\color{58c4dd} \text{number}} + {\color{58c4dd} \text{number}} + \cdots + {\color{58c4dd} \text{number}}}_{\text{{\color{83c167} count} times}} = {\color{ffee00} \text{total}}.

{\color{0000cc} \text{number}} \cdot {\color{008000} \text{count}} = \underbrace{{\color{0000cc} \text{number}} + {\color{0000cc} \text{number}} + \cdots + {\color{0000cc} \text{number}}}_{\text{{\color{008000} count} times}} = {\color{ff5500} \text{total}}.

The ${\color{83c167} \text{count}}$ ${\color{008000} \text{count}}$ tells us how many times to add the ${\color{58c4dd} \text{number}}$ ${\color{0000cc} \text{number}}$ to itself in order to get the ${\color{ffee00} \text{total}}\text{.}$ ${\color{ff5500} \text{total}}\text{.}$

As we take math classes, we're asked to solve multiplication problems. These problems present us with two of the three components and endeavor us with discerning the third. In elementary school, we're given the ${\color{58c4dd} \text{number}}$ ${\color{0000cc} \text{number}}$ and the ${\color{83c167} \text{count}}\text{,}$ ${\color{008000} \text{count}}\text{,}$ and we have to find the ${\color{ffee00} \text{total}}\text{.}$ ${\color{ff5500} \text{total}}\text{.}$ A question might be written as follows, where we have to fill in the box:

{\color{58c4dd} 6} \cdot {\color{83c167} 4} = {\color{ffee00} \fbox{\phantom{\;24y|}\!\!\!}}\,.

{\color{0000cc} 6} \cdot {\color{008000} 4} = {\color{ff5500} \fbox{\phantom{\;24y|}\!\!\!}}\,.

Our answer is the product ${\color{ffee00} 24}\text{:}$ ${\color{ff5500} 24}\text{:}$

{\color{58c4dd} 6} \cdot {\color{83c167} 4} = {\color{ffee00} \fbox{\;24\phantom{y|}\!\!\!}}\,.

{\color{0000cc} 6} \cdot {\color{008000} 4} = {\color{ff5500} \fbox{\;24\phantom{y|}\!\!\!}}\,.

Later on in our education, we're asked another type of question. Instead of the ${\color{58c4dd} \text{number}}$ ${\color{0000cc} \text{number}}$ and the ${\color{83c167} \text{count}}\text{,}$ ${\color{008000} \text{count}}\text{,}$ we're given the ${\color{58c4dd} \text{number}}$ ${\color{0000cc} \text{number}}$ and the ${\color{ffee00} \text{total}}\text{.}$ ${\color{ff5500} \text{total}}\text{.}$ We have to find the ${\color{83c167} \text{count}}$ ${\color{008000} \text{count}}$ and write it in the box:

\begin{aligned} {\color{58c4dd} 6} \cdot \,{\color{83c167} \fbox{\phantom{\,\;4y|\!\!}}}\, & = {\color{ffee00} 24}, \\ {\color{58c4dd} 6} \cdot \,{\color{83c167} \fbox{\,\;4\phantom{y|\!\!}}}\, & = {\color{ffee00} 24}. \end{aligned}

\begin{aligned} {\color{0000cc} 6} \cdot \,{\color{008000} \fbox{\phantom{\,\;4y|\!\!}}}\, & = {\color{ff5500} 24}, \\ {\color{0000cc} 6} \cdot \,{\color{008000} \fbox{\,\;4\phantom{y|\!\!}}}\, & = {\color{ff5500} 24}. \end{aligned}

Or, instead of a box, the question may use a variable:

{\color{58c4dd} 6}{\color{83c167} n} = {\color{ffee00} 24}.

{\color{0000cc} 6}{\color{008000} n} = {\color{ff5500} 24}.

Here we're asked to solve for ${\color{83c167} n}\text{.}$ ${\color{008000} n}\text{.}$ Which is just answering the question, how many times do we add ${\color{58c4dd} 6}$ ${\color{0000cc} 6}$ to get ${\color{ffee00} 24}\text{?}$ ${\color{ff5500} 24}\text{?}$ Of course we could figure this out by dividing both sides by ${\color{58c4dd} 6}\text{:}$ ${\color{0000cc} 6}\text{:}$

\begin{aligned} {\color{58c4dd} 6}{\color{83c167} n} & = {\color{ffee00} 24}, \\ \frac{\cancel{{\color{58c4dd} 6}}{\color{83c167} n}}{\cancel{{\color{58c4dd} 6}}} & = \frac{{\color{ffee00} 24}}{{\color{58c4dd} 6}}, \\ {\color{83c167} n} & = {\color{83c167} 4}. \end{aligned}

\begin{aligned} {\color{0000cc} 6}{\color{008000} n} & = {\color{ff5500} 24}, \\ \frac{\cancel{{\color{0000cc} 6}}{\color{008000} n}}{\cancel{{\color{0000cc} 6}}} & = \frac{{\color{ff5500} 24}}{{\color{0000cc} 6}}, \\ {\color{008000} n} & = {\color{008000} 4}. \end{aligned}

But let's think about how to solve this type of problem if we didn't know an algorithm for division. We could figure out the answer just by counting! We add ${\color{58c4dd} 6}$ ${\color{0000cc} 6}$ to itself until we get to ${\color{ffee00} 24}\text{,}$ ${\color{ff5500} 24}\text{,}$ then count the number of ${\color{58c4dd} 6}$ ${\color{0000cc} 6}$ 's we need:

\begin{aligned} & {\color{58c4dd} 6} = 6, \\ & {\color{58c4dd} 6} + {\color{58c4dd} 6} = 12, \\ & {\color{58c4dd} 6} + {\color{58c4dd} 6} + {\color{58c4dd} 6} = 18, \\ & \underbrace{{\color{58c4dd} 6} + {\color{58c4dd} 6} + {\color{58c4dd} 6} + {\color{58c4dd} 6}}_{{\color{83c167} 4}\text{ times}} = {\color{ffee00} 24}. \end{aligned}

\begin{aligned} & {\color{0000cc} 6} = 6, \\ & {\color{0000cc} 6} + {\color{0000cc} 6} = 12, \\ & {\color{0000cc} 6} + {\color{0000cc} 6} + {\color{0000cc} 6} = 18, \\ & \underbrace{{\color{0000cc} 6} + {\color{0000cc} 6} + {\color{0000cc} 6} + {\color{0000cc} 6}}_{{\color{008000} 4}\text{ times}} = {\color{ff5500} 24}. \end{aligned}

Our answer is ${\color{83c167} n = 4}\text{.}$ ${\color{008000} n = 4}\text{.}$

The algorithms we learn for long division and for finding and cancelling out common factors are definitely useful. But it's important to remember that they're tools we use for finding the ${\color{83c167} \text{count}}\text{.}$ ${\color{008000} \text{count}}\text{.}$ And if we wanted to, we could just do the counting ourselves.

Repeated Multiplication #

As we advance further in school, following up after repeated additions, we have repeated multiplications. And we represent these repetitions using exponents. For example,

{\color{58c4dd} 3}^{{\color{83c167} 5}} = \underbrace{{\color{58c4dd} 3} \cdot {\color{58c4dd} 3} \cdot {\color{58c4dd} 3} \cdot {\color{58c4dd} 3} \cdot {\color{58c4dd} 3}}_{{\color{83c167} 5}\text{ times}} = {\color{ffee00} 243}.

{\color{0000cc} 3}^{{\color{008000} 5}} = \underbrace{{\color{0000cc} 3} \cdot {\color{0000cc} 3} \cdot {\color{0000cc} 3} \cdot {\color{0000cc} 3} \cdot {\color{0000cc} 3}}_{{\color{008000} 5}\text{ times}} = {\color{ff5500} 243}.

Now we probably don't have ${\color{58c4dd} 3}^{{\color{83c167} 5}}$ ${\color{0000cc} 3}^{{\color{008000} 5}}$ memorized like we do our multiplication tables. But we can just multiply by ${\color{58c4dd} 3}$ ${\color{0000cc} 3}$ repeatedly to get the answer. We have the same three components as before: a ${\color{58c4dd} \text{number}}\text{,}$ ${\color{0000cc} \text{number}}\text{,}$ a ${\color{83c167} \text{count}}\text{,}$ ${\color{008000} \text{count}}\text{,}$ and a ${\color{ffee00} \text{total}}\text{.}$ ${\color{ff5500} \text{total}}\text{.}$ In general we can write

{\color{58c4dd} \text{number}}^{{\color{83c167} \text{count}}} = \underbrace{{\color{58c4dd} \text{number}} * {\color{58c4dd} \text{number}} * \cdots * {\color{58c4dd} \text{number}}}_{{\color{83c167} \text{count}}\text{ times}} = {\color{ffee00} \text{total}}.

{\color{0000cc} \text{number}}^{{\color{008000} \text{count}}} = \underbrace{{\color{0000cc} \text{number}} * {\color{0000cc} \text{number}} * \cdots * {\color{0000cc} \text{number}}}_{{\color{008000} \text{count}}\text{ times}} = {\color{ff5500} \text{total}}.

Here we use a $*$ instead of a $\cdot$ to denote multiplication, to avoid confusion with the center dots. In this context we call the ${\color{58c4dd} \text{number}}$ ${\color{0000cc} \text{number}}$ the base, the ${\color{83c167} \text{count}}$ ${\color{008000} \text{count}}$ the exponent, and the ${{\color{ffee00} \text{total}}}$ ${{\color{ff5500} \text{total}}}$ the product:

{\color{58c4dd} \text{base}}^{{\color{83c167} \text{exponent}}} = {\color{ffee00} \text{product}}.

{\color{0000cc} \text{base}}^{{\color{008000} \text{exponent}}} = {\color{ff5500} \text{product}}.

When you see the word exponent, think about the word expose. The term refers to the written appearance of the number as being exposed up and to the right, rather than to any particular mathematical property.

Just like with multiplication, math problems involving exponentiation give use two pieces of information and task us with finding the third. We start, once again, by being given a ${\color{58c4dd} \text{number}}$ ${\color{0000cc} \text{number}}$ and a ${\color{83c167} \text{count}}$ ${\color{008000} \text{count}}$ and being asked to find the ${\color{ffee00} \text{total}}\text{:}$ ${\color{ff5500} \text{total}}\text{:}$

{\color{58c4dd} 6}^{{\color{83c167} 3}} = {\color{ffee00} \fbox{\phantom{\;243y|}\!\!\!}}\,.

{\color{0000cc} 6}^{{\color{008000} 3}} = {\color{ff5500} \fbox{\phantom{\;243y|}\!\!\!}}\,.

And again, we can multiply ${\color{58c4dd} 6}$ ${\color{0000cc} 6}$ by itself ${\color{83c167} 3}$ ${\color{008000} 3}$ times to obtain the answer:

{\color{58c4dd} 6}^{{\color{83c167} 3}} = \underbrace{{\color{58c4dd} 6} \cdot {\color{58c4dd} 6} \cdot {\color{58c4dd} 6}}_{{\color{83c167} 3}\text{ times}} = {\color{ffee00} \fbox{\;243\phantom{y|}\!\!\!}}\,.

{\color{0000cc} 6}^{{\color{008000} 3}} = \underbrace{{\color{0000cc} 6} \cdot {\color{0000cc} 6} \cdot {\color{0000cc} 6}}_{{\color{008000} 3}\text{ times}} = {\color{ff5500} \fbox{\;243\phantom{y|}\!\!\!}}\,.

Now comes the more interesting question. Suppose, as we did previous with multiplication problems, that we're given a ${\color{58c4dd} \text{number}}$ ${\color{0000cc} \text{number}}$ and a ${\color{ffee00} \text{total}}\text{.}$ ${\color{ff5500} \text{total}}\text{.}$ How do we solve for the ${\color{58c4dd} \text{count}}\text{?}$ ${\color{0000cc} \text{count}}\text{?}$ Here we don't have the division algorithm to aide us like we did with our earlier analogous situation. But we still can resort to counting. Suppose we want to solve

{\color{58c4dd} 2}^{{\color{83c167} n}} = {\color{ffee00} 32}.

{\color{0000cc} 2}^{{\color{008000} n}} = {\color{ff5500} 32}.

We can do so by multiplying ${\color{58c4dd} 2}$ ${\color{0000cc} 2}$ by itself until we reach ${\color{ffee00} 32}$ ${\color{ff5500} 32}$ while counting the number of multiplications:

\begin{aligned} & {\color{58c4dd} 2} = 2, \\ & {\color{58c4dd} 2} \cdot {\color{58c4dd} 2} = 4, \\ & {\color{58c4dd} 2} \cdot {\color{58c4dd} 2} \cdot {\color{58c4dd} 2} = 8, \\ & {\color{58c4dd} 2} \cdot {\color{58c4dd} 2} \cdot {\color{58c4dd} 2} \cdot {\color{58c4dd} 2} = 16, \\ & \underbrace{{\color{58c4dd} 2} \cdot {\color{58c4dd} 2} \cdot {\color{58c4dd} 2} \cdot {\color{58c4dd} 2} \cdot {\color{58c4dd} 2}}_{{\color{83c167} 5}\text{ times}} = {\color{ffee00} 32}. \end{aligned}

\begin{aligned} & {\color{0000cc} 2} = 2, \\ & {\color{0000cc} 2} \cdot {\color{0000cc} 2} = 4, \\ & {\color{0000cc} 2} \cdot {\color{0000cc} 2} \cdot {\color{0000cc} 2} = 8, \\ & {\color{0000cc} 2} \cdot {\color{0000cc} 2} \cdot {\color{0000cc} 2} \cdot {\color{0000cc} 2} = 16, \\ & \underbrace{{\color{0000cc} 2} \cdot {\color{0000cc} 2} \cdot {\color{0000cc} 2} \cdot {\color{0000cc} 2} \cdot {\color{0000cc} 2}}_{{\color{008000} 5}\text{ times}} = {\color{ff5500} 32}. \end{aligned}

Thus our answer is ${\color{83c167} n = 5}\text{.}$ ${\color{008000} n = 5}\text{.}$ And just like we use the name division refer to the process of finding the ${\color{83c167} \text{count}}$ ${\color{008000} \text{count}}$ when working with repeated addition, we need a name for the concept of finding the ${\color{83c167} \text{count}}$ ${\color{008000} \text{count}}$ when working with repeated multiplication. And for that, it's time to introduce logarithms.

Defintion of a Logarithm #

Let's dive right in and state the definition.

Definition 1 (Logarithm). Let $x$ and $b$ be positive real numbers with $b$ not equal to $1\text{.}$ If $n$ is a number such that $b$ multiplied by itself $n$ times equals $x$ —that is, such that ${b^n = x}$ —we write

\log_b{x} = n.

This notation is read the base $b$ logarithm of $x$ equals $n\text{,}$ or more simply as log base $b$ of $x$ is $n\text{.}$ Here $b$ is called the base of the logarithm, and $x$ is called the argument of the logarithm.

We'll discuss more advanced properties and techniques for computing logarithms later. But for now let's just concentrate on what this definition is telling us.

When we have an exponential equation such as

{\color{58c4dd} 2}^{{\color{83c167} 5}} = {\color{ffee00} 32},

{\color{0000cc} 2}^{{\color{008000} 5}} = {\color{ff5500} 32},

we have another notation for writing it:

\log_{{\color{58c4dd} 2}}{{\color{ffee00} 32}} = {\color{83c167} 5}.

\log_{{\color{0000cc} 2}}{{\color{ff5500} 32}} = {\color{008000} 5}.

This equation is read, the base ${\color{58c4dd} 2}$ ${\color{0000cc} 2}$ logarithm of ${\color{ffee00} 32}$ ${\color{ff5500} 32}$ equals ${\color{83c167} 5}\text{.}$ ${\color{008000} 5}\text{.}$ It tells us that if we multiply ${\color{58c4dd} 2}$ ${\color{0000cc} 2}$ by itself ${\color{83c167} 5}$ ${\color{008000} 5}$ times, we get ${\color{ffee00} 32}\text{.}$ ${\color{ff5500} 32}\text{.}$

Now for some logarithm examples. If we're given an equation and asked to find what the exponent is, we can do what we did previous: multiply the base by itself until we reach the product, then count the number of times we multiplied.

${\log_{{\color{58c4dd} 2}}{{\color{ffee00} 64}}}\text{.}$ ${\log_{{\color{0000cc} 2}}{{\color{ff5500} 64}}}\text{.}$

To calculate this logarithm, write out products of $2$ until we reach $64\text{.}$

Hence ${\log_{{\color{58c4dd} 2}}{{\color{ffee00} 64}} = {\color{83c167} 6}}\text{.}$ ${\log_{{\color{0000cc} 2}}{{\color{ff5500} 64}} = {\color{008000} 6}}\text{.}$

Some Caveats About the Definition #

In our definition of $\log_b{x}\text{,}$ we impose restrictions on the numbers $b$ and $x\text{.}$ We require that both of these are positive, and further that $b$ is not equal to $1\text{.}$ What do these conditions mean, and why are they important?

We define the base $b$ logarithm of $x$ only when the the following conditions hold:

For example, what happens if $b$ or $x$ is zero? Well, $\log_2{0}$ is not defined, since there is no power of $2$ such that ${2^n = 0}\text{.}$ Multiplying $2$ by itself just gives bigger and bigger positive numbers, we can't multiply by $2$ and get zero. Similarly, since ${0 \cdot 0 = 0}\text{,}$ there is no $n$ such that ${0^n = 2}\text{.}$ Multiplying repeatedly by zero keeps yielding zero. Hence $\log_0{2}$ is not defined either.

You'll find that we don't gain any utility from attempting to include other cases in our definition of logarithms, anyway. In a more advanced course we could study how complex numbers relate to logarithms. But for now, and for general-purpose usage of logarithms with real numbers, these restrictions suffice.