When it comes to the relationship $a^b=c$, we have examined how $c$ can be thought of as a combination of $a$ and $b$. We have also considered the implications of thinking of $a$ as a combination of $b$ and $c$. There is one more possible combination we could contemplate -- what happens if we view $b$ as a combination of $a$ and $c$?

John Napier considered something similar in 1614, calling such combinations "logarithms" (a term he invented by jamming together the Greek words for "proportion" (i.e., *logos*) and "number" (i.e., *arithmos*).

In modern mathematics, we abbreviate the value resulting from the combination of $a$ and $c$ that will yield $b$ above with $\log_a c$. (*Note, this presumes $a \ne 1$ is positive, as otherwise there will be problems.*) That is to say:
$\newcommand{\dotriangle}[1]{\raise{-0.7ex}{\vcenter{#1 \kern .2ex\hbox{$\triangle$}\kern.2ex}}}$
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$\newcommand{\vtp}[3]{\vcenter{\tripow{#1}{#2}{#3}}}$
$$a^b = c \quad \Longleftrightarrow \quad \log_a c = b$$

Recall we have previously referred to the $a$ in the equation on the left as the "base" of the power $a^b$. In a similar way, we call the $a$ in the equation on the right the **base** of the logarithm $\log_a c$, which we then read as "the log base $a$ of $c$".

*Often in the sciences and elsewhere, folks must deal with very large or very small values and employ scientific notation, which uses powers with a base of $10$, to express these. As such, it is not uncommon for texts to adopt the convention of writing $\log x$ when they really mean $\log_{10} x$. However, one must be careful! In computer science, where data is represented in "bits" which take on two states, powers with a base of $2$ are more common. In texts in this discipline, one may find $\log x$ meant to be $\log_2 x$. In still other disciplines, yet another value (Euler's number, $e$) will be more "natural" in calculations, and $\log x$ will be used with this base to be implied instead. The upshot of this is that when the base is not written, it is assumed to be the base most common to the discipline in question. As such, when we write $\log x$, we say we have taken the common log of $x$.*

We can -- and will -- explore the properties associated with these logarithms. However, before doing so, let us pause for a moment and consider the beautiful symmetry connecting these three ideas: powers, roots, and logarithms

... a symmetry some have enshrined in a new notation known as the "**Triangle of Power**".

As far as notations go, the "triangle of power" is admittedly so young that one could conceive of it as still gestating in the womb! It made its first appearance on Stack Exchange in July of 2012, but most who know of it now do so due to Grant Sanderson endorsing it four years later on his YouTube channel 3Blue1Brown which is dedicated to visualizing mathematical ideas through beautiful animations -- a channel that boasts 4.51 million subscribers as of 2022, by the way!

As new and as-yet-unaccepted by the general mathematics community as it still is -- once seen -- many will agree it at least meets the criteria for good notation established in the following quotation by famed British logician Bertrand Russell:

"A good notation has a subtlety and suggestiveness which at times make it almost seem like a live teacher. -- B. Russell"

Here's how triangle of power notation works: First, we express the relationship $a^b = c$ by writing: $\tripow{a}{b}{c}$

This triangle then serves as a basis for denoting each of the three values at its vertices as a combination of the other two. We do this by duplicating the triangle above, but leaving the value the combination equals missing.

Consider how much cleaner and more symmetric the results appear (see the statements below on the left) as compared to the same facts written with more traditional notation (on the right): $$\begin{array}{cclcrcl} \vcenter{\tripow{a}{b}{}} \mkern -0.4em&=& \mkern -0.5em c \mkern 1em & \quad \longleftrightarrow \mkern 0.4em \quad & a^b \mkern -0.6em &=& \mkern -0.5em c\\ \vcenter{\tripow{a}{}{c}} \mkern -0.8em&=& \mkern -0.5em b & \quad \longleftrightarrow \mkern 0.4em \quad & \log_a c \mkern -0.6em &=& \mkern -0.5em b\\ \vcenter{\tripow{}{b}{c}} \mkern -1.2em&=& \mkern -0.5em a & \quad \longleftrightarrow \mkern 0.4em \quad & \sqrt[b]{c} \mkern -0.6em &=& \mkern -0.5em a \end{array}$$

So that we can manipulate such expressions without first "translating back" to our traditional notations, note how some of our familiar rules for exponents look in "triangle of power" notation:

$$\begin{array}{rcl} x^m \cdot x^n = x^{m+n} &\quad \longleftrightarrow \quad& \vcenter{\tripow{x}{m}{} \cdot \tripow{x}{n}{} = \tripow{x}{m+n}{}}\\\\ \cfrac{x^m}{x^n} = x^{m-n} &\quad \longleftrightarrow \quad& \vcenter{\cfrac{\tripow{x}{m}{}}{\tripow{x}{n}{}}} = \vcenter{\tripow{x}{m-n}{}}\\\\ (xy)^n = x^n y^n &\quad \longleftrightarrow \quad& \vcenter{\tripow{xy}{n}{} = \tripow{x}{n}{} \cdot \tripow{y}{n}{}}\\\\ \displaystyle{\left(\frac{x}{y}\right)^n} = \cfrac{x^n}{y^n} &\quad \longleftrightarrow \quad& \vcenter{\displaystyle{\tripow{\textstyle \frac{x}{y}}{n}{}} = \cfrac{\tripow{x}{n}{}}{\tripow{y}{n}{}}}\\\\ (x^m)^n = x^{mn} &\quad \longleftrightarrow \quad& \vcenter{\displaystyle{\tripow{\tripow{x}{m}{}}{n}{}} = \textstyle \tripow{x}{mn}{}}\\\\ \end{array}$$Admittedly, the new notation doesn't really seem to look any better for exponents. Indeed, it actually looks more cumbersome here, in that we are using more "ink" to express the same ideas. That said, notice how the following rules for radicals (for appropriate values of $x$, $y$, $m$, and $n$) compare:

$$\begin{array}{rcl} \sqrt[n]{x} \cdot \sqrt[n]{y} = \sqrt[n]{xy} &\quad \longleftrightarrow \quad& \vcenter{ \tripow{}{n}{xy} = \tripow{}{n}{x} \cdot \tripow{}{n}{y}}\\\\ \displaystyle{\sqrt[n]{\frac{x}{y}}} = \cfrac{\sqrt[n]{x}}{\sqrt[n]{y}} &\quad \longleftrightarrow \quad& \vcenter{\displaystyle{\tripow{}{n}{\textstyle \frac{x}{y}}} = \cfrac{\tripow{}{n}{x}}{\tripow{}{n}{y}}}\\\\ \displaystyle{\sqrt[n]{\sqrt[m]{x}^{\phantom{1}}}} = \sqrt[m n]{x^{\phantom{1}}} &\quad \longleftrightarrow \quad& \vcenter{\displaystyle{\tripow{}{n}{\tripow{}{m}{x}}} = \textstyle \tripow{}{mn}{x}}\\\\ \end{array}$$Here we start to see the niceties of this notation -- look how symmetric the 3 results above look in comparison to the last 3 results we gave for exponents!

The symmetry doesn't stop there. Recall how exponents, roots, and logarithms can sometimes undo one another: When $a$ and $b$ are positive, we have

$$\begin{array}{rcl} \sqrt[a]{b^a} = b &\quad \leftrightarrow \quad& \displaystyle{\tripow{}{a}{\tripow{b}{a}{}}} = b\\\\ (\sqrt[a]{b})^a = b &\quad \leftrightarrow \quad& \displaystyle{\tripow{\tripow{}{a}{b}}{a}{} = b}\\\\ a^{\log_a b} = b &\quad \leftrightarrow \quad& \displaystyle{\tripow{a}{\tripow{a}{}{b}}{} = b}\\\\ \end{array}$$(In case the last relation wasn't obvious, recall $a^x = b \Longleftrightarrow \log_a b = x$, and then substitute the $x$ in the first with the $\log_a b$ of the second.)

Looking at the structure of the triangle notation seen in the last three examples, notice how each large triangle has an $a$ at one vertex with the smaller triangle having an $a$ and $b$ in the two positions closest to the big triangle. Then, in each case we can "cancel" both values of $a$.

Kinda' makes one wonder what happens when we see this same structure in the other ways it could occur, right? Wouldn't it just "feel right" if the following were all equal to $b$ too? $$\tripow{a}{}{\tripow{a}{b}{}} = b \quad \quad \quad \tripow{\tripow{}{b}{a}}{}{a} = b \quad \quad \quad \tripow{}{\tripow{b}{}{a}}{a} = b$$

But this would mean all the following are true, respectively... $$\log_a a^b = b \quad \quad \quad \quad \log_{\sqrt[b]{a}} a = b \quad \quad \quad \quad \sqrt[(\log_b a)]{a} = b$$

They are!

The first of the three above (as well as the last of the three before these) is traditionally taught to students as one of the "inverse relationships" for logs and is normally justified like this: We know $a^b = c$ is equivalent to $\log_a c = b$. In this way, we can think of $\log_a c$ as "*the exponent needed on $a$ to produce a power equal to $c$*". As such, we can interpret $\log_a a^b$ as "*the exponent needed on $a$ to produce $a^b$*", which is of course $b$.

Likewise, the earlier seen $a^{\log_a b}$ can be interpreted as "*the value $a$ raised to precisely the power needed to raise $a$ to get $b$*". Consequently, we not surprisingly get $b$.

The last two of the three equations above are inverse relationships not as often seen -- but every bit as valid! Note, the reader will benefit from taking a moment to convince themselves these two equations are also true.

The "triangle of power" notation has more to tell us about logarithms, revealing even more symmetries...

Suppose $\tripow{x}{m}{a}$ and $\tripow{x}{n}{b}$ for positive $x$, $a$, and $b$. (That is to say, suppose $x^m = a$ and $x^n = b$.)

As a consequence,

But then, $$\begin{array}{rcll} \tripow{x}{}{ab} &=& \displaystyle{\tripow{x}{}{\tripow{x}{m+n}{}}} &\quad {\scriptstyle \textrm{after substituting } \tripow{x}{m+n}{} \textrm{ for } ab}\\\\ &=& m+n &\quad {\scriptstyle \textrm{after "canceling" the } x \textrm{, as discussed earlier}}\\\\ &=& \vcenter{\tripow{x}{}{a}} + \vcenter{\tripow{x}{}{b}} &\quad {\scriptstyle \textrm{after rewriting $m$ and $n$ in triangle of power notation}} \end{array}$$

As such, we have

In traditional notation, we just proved the perhaps more familiar result that for positive $x$, $a$ and $b$:

Again, notice the symmetry between this and the exponent rule for products of powers of the same base:: $$\begin{array}{rcl} \vcenter{\tripow{x}{}{ab}} = \vcenter{\tripow{x}{}{a}} + \vcenter{\tripow{x}{}{b}} &\quad \longleftrightarrow \quad& \log_x ab = \log_x a + \log_x b\\\\ \vcenter{\tripow{x}{a+b}{\phantom{ab}}} = \vcenter{\tripow{x}{a}{\phantom{a}}} \cdot \vcenter{\tripow{x}{b}{\phantom{b}}} &\quad \longleftrightarrow \quad& x^{a+b} = x^a \cdot x^b \end{array}$$

A similar argument using "triangle of power" notation establishes the following logarithmic relationship (i.e., the log of a quotient is a difference of logs).

Again suppose $\vtp{x}{m}{a}$ and $\vtp{x}{n}{b}$ for positive $x$, $a$, and $b$. Then, $$\textstyle{\cfrac{a}{b} = \cfrac{\vtp{x}{m}{}}{\vtp{x}{n}{}} = \vtp{x}{m-n}{}}$$ But then $$\begin{array}{rcl} \vtp{x}{}{\frac{a}{b}} &=& \dtp{x}{}{\vtp{x}{m-n}{}}\\ &=& m-n\\ &=& \vtp{x}{}{a} - \vtp{x}{}{b} \end{array}$$ As such, we have $$\textstyle{\vtp{x}{}{\frac{a}{b}} = \vtp{x}{}{a} - \vtp{x}{}{b}}$$ In traditional notation, we just proved the following important result for positive $x$, $a$, and $b$: $$\log_x \frac{a}{b} = \log_x a - \log_x b \quad \textrm{(i.e., the log of a quotient is a difference of logs)}$$

Note the symmetry this logarithmic rule has with the rule of exponents that directs us to subtract exponents when dividing powers of the same base, as shown: $$\begin{array}{rcl} \vcenter{\tripow{x}{}{a \div b}} = \vcenter{\tripow{x}{}{a\vphantom{\frac{a}{b}}}} - \vcenter{\tripow{x}{}{b\vphantom{\frac{a}{b}}}} &\quad \longleftrightarrow \quad& \log_x \frac{a}{b} = \log_x a - \log_x b\\\\ \vcenter{\tripow{x}{a-b}{\phantom{a \div b}}} = \vcenter{\tripow{x}{a}{\phantom{a}}} \div \vcenter{\tripow{x}{b}{\phantom{b}}} &\quad \longleftrightarrow \quad& x^{a-b} = \cfrac{x^a}{x^b} \end{array}$$

Of course, once we have $\vcenter{\tripow{x}{}{ab}} = \vcenter{\tripow{x}{}{a}} + \vcenter{\tripow{x}{}{b}}$, the suggestion that $\vtp{x}{}{a^n} = n \left(\vtp{x}{}{a}\right)$ is immediate given the obvious pattern appearing below: $$\begin{array}{rcl} \vtp{x}{}{a^2} &=& \vtp{x}{}{a} + \vtp{x}{}{a\phantom{1}} = \vtp{x}{}{a} + 1 \left(\vtp{x}{}{a}\right) = 2 \left(\vtp{x}{}{a}\right)\\ \vtp{x}{}{a^3} &=& \vtp{x}{}{a} + \vtp{x}{}{a^2} = \vtp{x}{}{a} + 2 \left(\vtp{x}{}{a}\right) = 3\left(\vtp{x}{}{a}\right)\\ \vtp{x}{}{a^4} &=& \vtp{x}{}{a} + \vtp{x}{}{a^3} = \vtp{x}{}{a} + 3 \left(\vtp{x}{}{a}\right) = 4\left(\vtp{x}{}{a}\right)\\ & \vdots & \end{array}$$

While the pattern above leads naturally to a proof (by induction^{†}) of the suggested result, such an argument would only establish it for positive integers $n$. However, we can do better.

*† If you are curious what this "induction" thing is all about, check out the following notes on the subject: The Principle of Mathematical Induction and Variations on Induction. We don't need to know about this important means of mathematical argument just yet -- but we will eventually.*

The following more direct (and shorter) argument will establish the same, but for all real values $n$:

Suppose $y = \vtp{x}{}{a}$. Then $x^y = a$. Raising both sides to the $n^{th}$ power, we have $(x^y)^n = a^n$. But then, $x^{ny} = a^n$, so $\vtp{x}{}{a^n} = ny$. Finally, swapping out $y$ with $\vtp{x}{}{a}$ gives us the desired result: $$\textstyle{\vtp{x}{}{a^n} = n \left(\vtp{x}{}{a}\right)}$$

Of course, we must acknowledge the calculations immediately above actually "mix" traditional and triangle-of-power notations. For the triangle-of-power purists out there, note that we could also write our result in this form: $$\displaystyle{\dtp{x}{}{\tripow{a}{n}{}} = n \left(\tripow{x}{}{a}\right)}$$

Still, it is perhaps easier to see the parallels between this result and the traditionally-written "log-of-a-power" rule for logarithms if we use the aforementioned "mixed form":

Another important theorem for logarithms also follows quickly from the exponent rule whereby $(a^x)^y = a^{xy}$. To see this, consider the following:

Suppose $\tripow{a}{x}{b}$ and $\tripow{b}{y}{c}$ for positive $a$, $b$, and $c$ (that is, $a^x=b$ and $b^y=c$)

But then, $$\begin{array}{rcl} c &=& \tripow{b}{y}{}\\\\ &=& \displaystyle{\vtp{\tripow{a}{x}{}}{y}{}} \quad {\scriptstyle \textrm{ after substituting $\vtp{a}{x}{}$ for $b$}}\\\\ &=& \tripow{a}{xy}{} \quad {\scriptstyle \textrm{ after applying the "triangle of power" variant of $(a^x)^y = a^{xy}$}}\\\\ \end{array}$$

The conclusion can be written in multiple ways:Focusing on the middle one, it must be that

$$\begin{array}{rcl} \vcenter{\tripow{a}{}{c}} &=& xy\\ &=& \vcenter{\tripow{a}{}{b}} \cdot \vcenter{\tripow{b}{}{c}}\\\\ \end{array}$$ Dividing both sides by $\vcenter{\tripow{a}{}{b}}$ yields $$\cfrac{\tripow{a}{}{c}}{\tripow{a}{}{b}} = \vcenter{\textstyle \tripow{b}{}{c}}$$ or again, in the more traditional notation -- the "Change of Bases" rule for logarithms: $$\cfrac{\log_a c}{\log_a b} = \log_b c$$The theorem bears this name as it can be used to change the base of any and all logarithms that might appear in an expression to any desired legal value.

This is particularly important when one needs to evaluate a logarithm on a calculator. Depending on the sophistication of one's calculator, you might only have a single "`log`

" button that finds $\log x$ with an assumed "common base" of $10$.

Armed with only this, what if you need to find the value of $\log_2 7 + \log_3 5$? Do you need a fancier calculator? Certainly not!

With the change of bases theorem as stated above, and using $a=10$, we can rewrite the value we seek: $$\log_2 7 + \log_3 5 = \frac{\log_{10} 7}{\log_{10} 2} + \frac{\log_{10} 5}{\log_{10} 3}$$ which could then be typed into the calculator as

`log(7)/log(2) + log(5)/log(3)`

Using the traditional notation, logarithms are sometimes written with parentheses and sometimes without. For example, the following are equivalent $$\log_2 8 = \log_2 (8)$$ When the value of the "power" involved (e.g., the $8$ above) is some explicit number, the form on the left is often preferred (as it uses less ink). However, consider the expression $\log_2 4 + 4$. Should this be thought of as $\log_2 (4+4)$ or $(\log_2 4) + 4$? As the first equals $3$ and the second equals $6$, it clearly makes a difference! In such cases one should use parentheses to clear up any ambiguity.

Curiously though, we can make an argument that there is no real confusion when writing something like $\log_{7} ab$. In particular, we can claim $\log_7 ab$ unambigously means $\log_7 (ab)$.

This conclusion can be drawn from a couple of things.

First, the $b$ is simply drawn much closer to the $a$ than the $\log_7$ is -- suggesting by their proximity a "grouping" together of these two factors, which then results in prioritizing their combination before others. Second, when we write mathematical expressions (and when we have commutativity of our "products"), we generally write smaller, simpler factors before larger, more complicated ones.

For example, consider the two products $$2xy(x+6)\sqrt{x^5-3x+9} \quad \textrm{ vs. } \quad \sqrt{x^5-3x+9}xy(x+6)2$$ For a given value of $x$ and $y$, these two expressions absolutely calculate the same value. However, the one on the left -- given its "simple-to-more-complicated" ordering of its factors has some advantages.

First, note that if one writes the second expression out by hand, it is really easy to accidentally overextend the veniculum on the radical (i.e., the bar at the top) to a point where it appears the $x$ (or maybe even both $x$ and $y$ depending on how sloppy one's handwriting might be) were *inside* the radical! This of course, would be a very different expression!

Similarly, if we wrote the $2$ on the end of the second expression a bit higher than intended, it might be erroneously interpreted as squaring something instead of doubling something.

The first expression avoids these two pitfalls.

Therefore, having a default of writing things in a "simple-to-complex" form as we move from left to right in an expression is a good idea (*as long as doing so is legal -- remember a lack of commutativity would cause problems here*) -- and is a general convention followed by mathematicians. The reader is strongly encouraged to adopt this convention too!

As such, thinking of $\log_2 xy$ as $(\log_2 x) \cdot y$ would be awkward! The $y$ factor is far simpler/smaller than the $\log_2 x$ factor -- it should come first! Given that we didn't write it that way however, tells us that's not the way to interpret the expression. We should therefore interpret $\log_2 xy$ as $\log_2 (xy)$ instead.

Of course, if we play this game of saving a few drops of ink by leaving off parentheses -- there are other costs as well. For example, suppose you wanted to square $\log_3 x$. Writing $\log_3 x^2$ is definitely *not* what we want to write in this case, as the whole "simple-to-complex" discussion above would suggest this be read as $\log_3 (x^2)$.

We *could* write $(\log_3 x)^2$. That would be interpreted clearly as the square of the logarithm. Indeed, using parentheses almost always makes our intentions more clear.

However, there is a third alternative! For those that wish to "have one's cake and eat it too" -- we could write $\log_3^2 x$. Notice this avoids the parentheses but also can't be mistaken for the log of a square, when we want a square of a log instead.

This third alternative is a common one you will see -- and not only when it comes to logarithms, but in other contexts too. Just be explicit, using the following notation for powers of logarithms is perfectly acceptable: $$\log_b^n x = (\log_b x)^n$$

Recall that irrational values are simply values that are *not* rational. That is to say, they are values that are *not* writable as a ratio/fraction of two integers.

You might have been told at some point that irrational values when written as decimal values don't terminate or repeat. By "don't terminate", we mean we can't write them with a finite number of digits. By "don't repeat", we mean there is no sequence of digits in the decimal expansion that (at some point) simply occurs over and over again, without end.

Note that the value $1.2373496496496\cdots$, starting with the fifth decimal digit appears to repeat the sequence of digits "$496$" over and over. The suggestion is that because of this (presumed infinite) repetition, the resulting value must be rational.$\require{color}$

Putting off for the moment how we might know that, we must acknowledge that without complete knowledge of *all* of the digits past some given point in a number, we don't really know if there really is the *infinite* repetition we say we need, or if there is just some initial repetition of some digit sequence some number of times that eventually breaks down. The "$\cdots$" present above only means there are more digits past the last digit shown. The number $1.23734496496496\cdots$ could easily be $1.2373496496496{\color{red}859}\cdots$ when presented with more accuracy!

There is a special notation however, that we can use to let the reader know that we really do mean the value where a given sequence of digits keeps repeating from a given point in its decimal expansion onwards. In this notation, we simply draw a line over the repeating sequence. So for example,

Now let us return to the suggestion that numbers that don't repeat in this way -- numbers that *can't* be written in this "overline" notation -- must be irrational. How do we know that to be true?

Surprisingly, the answer is a simple one: For any number that *can* be written in this overline notation, there is an easy method/algorithm for rewriting these as fractions of integers (i.e., rational values).

Consider the following demonstration of this algorithm on the number $1.2373\overline{496}$ mentioned above:

First, let $x = 1.2373\overline{496}$

Let $d$ be the number of digits in the repeating sequence. Here $d=3$. Then, multiply both sides of the equation above by $10$ exactly $d$ times. This results in $1000x = 1237.3496\overline{496}$, in this case.

Subtract the first equation from the second. That is to say, note that: $$\begin{array}{rcl} 1000x &=& 1237.3496\overline{496}\\ x &=& \phantom{123}1.2373\overline{496}\\\hline 999x &=& 1236.1123 \end{array}$$

Multiply both sides by $10$ enough times to eliminate any digits to the right of the decimal point, if needed. Here, we need to multiply by $10$ four times to find $$9990000x = 12361123$$

Finally, divide both sides by the number to the left of $x$ to find $$x = \frac{12361123}{9990000}$$

As we have shown $x = 1237.3\overline{496}$ is representable as a fraction of integers, $x$ must be rational. A similar argument can be constructed for *any* decimal sequence that has an infinitely repeating digit sequence in this way. Thus, irrational values must necessarily have digits that *don't repeat*.

The argument that irrational values' decimal expansions don't terminate (i.e., they have a finite number of digits) is even easier to show. Again, consider the following example. Suppose $y= 1.287$, a terminating decimal expansion with only $3$ digits to the right of the decimal. We first multiply both sides by $10$ enough times to clear the decimal values. Here, that gives us $1000y = 1287$. Then, we divide both sides by the value to the left of the $y$ to reveal $$y = \frac{1287}{1000}$$ Again, seeing a fraction for this (and any other terminating decimal expansion), we know such values are rational. Thus, for a value to irrational requires its decimal expansion not terminate.

*As an interesting addendum to the above, the method given for turning a repeating decimal into a fraction can also be used to establish that $0.999\overline{9} = 9/9 = 1$. This means that $0.999\overline{9}$ and $1$ are not just really, really close to one another -- they are the exact same value!*

Playing around with the `log`

button on your calculator a bit might lead one to wonder if a lot of logarithm values are irrational. They often have "long and messy" looking decimal values that don't seem to terminate or contain any sequence of digits simply repeated over and over.

As has been said previously however, if you only have access to the first 14 digits or so of the values in question (calculators can only show you so many digits, after all) -- there is no way to tell if the value terminates or starts repeating some time after that!

However, there is another way we could argue that some logarithm values are irrational...

Most will believe from their experience^{‡} that prime factorizations of positive integers are unique (i.e., that is to say, there is only one way to break down any positive integer into a product of primes, up to the order of the factors in question).

For example, $120 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5$. We could reorder the factors given the commutative property of real numbers under multiplication -- but that's not telling us anything new. As such, re-ordering the factors does not result in a new prime factorization of a number. The important part of this is that there will *never* be a different prime factorization of $120$. When $120$ is broken down into a product of primes, there will *always* be exactly $3$ twos, $1$ three, and $1$ five.

*‡ : As with most claims in mathematics, we don't have to just rely on our experiences and previous observations to tell us this -- we can prove that prime factorizations are unique. We won't do that here as it takes us too far afield from our content of interest -- but interested students might look up the celebrated "Fundamental Theorem of Arithmetic" in textbook on "Number Theory".*

Now, consider the value $x = \log_2 3$. Suppose the value *was* rational. Then, we could write $x$ as a ratio/fraction of two integers, right? In other words, we could find integers $p$ and $q$ so that
$$\log_2 3 = \frac{p}{q}$$
If this is true however, consider what happens when we multiply both sides by $q$ and then apply the logarithm rule that involves "logs of powers and multiples of logs":
$$q \log_2 3 = p \quad \longrightarrow \quad \log_2 3^q = p$$
Now convert the equation on the right to exponential form:
$$2^p = 3^q$$
Think about what this says, however. Suppose $y$ is the common (positive integer) value shared by $2^p$ and $3^q$ above.

Given that $y = 2^p$ and noting $2$ is prime, the prime factorization of $y$ must be $\underbrace{2 \cdot 2 \cdots 2}_{p \textrm{ times}}$

But $y=3^q$ and $3$ is prime as well -- so we know the prime factorization of $y$ must (*also?*) be $\underbrace{3 \cdot 3 \cdots 3}_{q \textrm{ times}}$

Given what we know about prime factorizations (i.e., one never has more than one for any given positive integer), this is clearly impossible! We have run into a contradiction!

Thus, our original assumption that $\log_2 3$ was rational must have been incorrect.

Clearly, $\log_2 3$ must be irrational!

Nice, right?

We can numerically approximate logarithm values (without using the `log`

button on our calculator) by taking advantage of the various properties of logarithms mentioned earlier in this section.

As an example, suppose we wish to approximate $\log 7532$ (presuming the "common log" in this instance involves a base of $10$):

It will be useful to have approximations for a few other common log values to this end. Although some sets of values are more useful than others.

Since we just discussed prime factorizations and $7532 = 2^2 \cdot 7 \cdot 269$, suppose we knew $\log 2 \doteq 0.3010$, $\log 7 \doteq 0.8451$, and $\log 269 \doteq 2.4298$.

Then $$\begin{array}{rcl} \log 7532 &=& \log (2^2 \cdot 7 \cdot 269)\\ &=& \log 2^2 + \log 7 + \log 269\\ &=& 2 \log 2 + \log 7 + \log 269\\ &=& 2(0.3010) + (0.8451) + (2.4298)\\ &=& 3.8769 \end{array}$$

That said, one might wonder how we find the initial approximations for $\log 2$, $\log 7$, and $\log 269$, since the numbers involved are prime and don't break down any further?

To address these, let us consider the following related (but different) technique for approximating a logarithm:

Suppose we again want to find $\log 7532$. We should note right away that the value we seek will be between $3$ and $4$ as $10^3 \le 7532 \le 10^4$. To see this, note that every integer with $d$ digits will be between $10^{d-1}$ and $10^d$.

Thus, $\log 7532 = \log (10^3 \cdot 7.532) = 3 + \log 7.532$.

It remains to find $\log 7.532$.

Recall that we have a way (i.e., the Babylonian method) to approximate square roots. Noting that a square root of a square root is a fourth root, and a square root of a fourth root is an eighth root, etc -- we have a way to approximate $\sqrt{10}$, $\sqrt[4]{10}$, $\sqrt[8]{10}$, etc. Written with rational exponents instead, this means we can approximate $10^{1/2}$, $10^{1/4}$, $10^{1/8}$, etc.

Avoiding the tedium of these calculations, let us simply list the values that result: $$\begin{array}{rcl} 10^{1/2} &\doteq& 3.16227766\\ 10^{1/4} &\doteq& 1.77827941\\ 10^{1/8} &\doteq& 1.333521432\\ 10^{1/16} &\doteq& 1.154781985\\ 10^{1/32} &\doteq& 1.074607828\\ 10^{1/64} &\doteq& 1.036632928\\ 10^{1/128} &\doteq& 1.018151722\\ 10^{1/256} &\doteq& 1.009035045\\ 10^{1/512} &\doteq& 1.004507364\\ 10^{1/1024} &\doteq& 1.002251148\\ \vdots & & \end{array}$$

Now, see what happens when you divide the $7.532$ we have left by the $10^{1/2} \doteq 3.16227766$ found above. If the result is $\ge 1$, write down $10^{1/2}$ and update "what is left" to be this new quotient. If not, try the next power of $10$ in our list (i.e., $10^{1/4} \doteq 1.77827941$) and do the same. Here, the $7.532$ we had left gets updated to become $2.81827534$.

In the rare case that we get $0$ left, we stop. However, more often we will simply continue -- i.e., we look for the next largest power $10^{1/2^k}$ we can divide what's left by to stay over $1$, updating what's left to be this quotient, and then doing this over and over.

Here, we discover we can divide our initial "left over" value of $7.532$ first by $10^{1/2}$, and then by $10^{1/4}$ and $10^{1/8}$, while always keeping the quotient greater than $1$. However, division by an additional $10^{1/16}$ or $10^{1/32}$, or anything bigger than $10^{1/512}$ drops the quotient less than $1$ -- so we don't do any of these divisions. Instead, we divide by $10^{1/1024}$.

The more divisions we make, the better our approximation will be -- however, let's see how to put just what we have together.

Note that we can now say $10^3 \cdot 10^{1/2} \cdot 10^{1/4} \cdot 10^{1/8} \cdot 10^{1/1024}$
is a slight underestimate of $7532$. (*We know this product is less than $7532$ as we never let our aforementioned quotients get below $1$.*)

As such, the (common) log of this big product will be a slight underestimate of $\log 7532$, and $$\begin{array}{rcl} \log 7532 &\doteq& \log (10^3 \cdot 10^{1/2} \cdot 10^{1/4} \cdot 10^{1/8} \cdot 10^{1/1024})\\ &\doteq& \log 10^3 + \log 10^{1/2} + \log 10^{1/4} + \log 10^{1/8} + \log 10^{1/1024}\\ &\doteq& 3 + 1/2 + 1/4 + 1/8 + 1/1024\\ &=& 3.8759765625 \end{array}$$ Had we done more divisions than what we did above, we would have an even better approximation of $$3 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{1024} + \frac{1}{2048} + \frac{1}{4096} + \frac{1}{8192} + \frac{1}{16384} + \frac{1}{65536} + \frac{1}{524288}$$ which is just a bit larger than $3.8766$.

Comparing this to what your calculator's `log`

button produces (i.e., $3.87691031$), we see we did pretty good!

There are methods that will converge on the actual value of the logarithm in question much more quickly -- but these will require we learn a bit more first. As such, let us not worry about these (..yet! 😀)