# Alternative notation for exponents, logs and roots?

If we have

$$x^y = z$$

then we know that

$$\sqrt[y]{z} = x$$

and

$$\log_x{z} = y .$$

As a visually-oriented person I have often been dismayed that the symbols for these three operators look nothing like one another, even though they all tell us something about the same relationship between three values.

Has anybody ever proposed a new notation that unifies the visual representation of exponents, roots, and logs to make the relationship between them more clear? If you don't know of such a proposal, feel free to answer with your own idea.

This question is out of pure curiosity and has no practical purpose, although I do think (just IMHO) that a "unified" notation would make these concepts easier to teach.

• (BTW, is it just me or is the TeX preview on the question form not working today?) – friedo Mar 31 '11 at 1:48
• My TeX preview hasn't been working either. I don't understand your question though, what is wrong with the current notation? – Eric Naslund Mar 31 '11 at 1:55
• There's nothing wrong with it, I just think it's inelegant to have three symbols that are so different to describe three parts of the same relationship. I think it would be helpful for learners to see the relationship between logs and roots visually. – friedo Mar 31 '11 at 2:09
• Be careful about saying that these three statements are equivalent when the corresponding functions aren't always well-defined for all $x, y, z$... restricting to positive reals makes everything okay, though. – Qiaochu Yuan Jul 3 '12 at 20:42
• This whole program is totally misguided. There are three different symbols because there are three qualitatively different functions. To have analogous notation for the logarithmic and exponential functions - e.g. by using a triangle with 3 seemingly symmetric vertices - would be as actively harmful as to have similar words for "giving the birth" and "murdering". Also, the natural elementary functions are just ln(x) and exp(x) which only have one argument, not two, and the triangle-style notation further prevents people from understanding why e=2.718... is the most natural base. – Luboš Motl Jul 19 '16 at 5:01

Always assuming $x>0$ and $z>0$, how about: \begin{align} x^y &={} \stackrel{y}{_x\triangle_{\phantom{z}}}&&\text{x to the y}\\ \sqrt[y]{z} &={} \stackrel{y}{_\phantom{x}\triangle_{z}}&&\text{yth root of z}\\ \log_x(z)&={} \stackrel{}{_x\triangle_{z}}&&\text{log base x of z}\\ \end{align} The equation $x^y=z$ is sort of like the complete triangle $\stackrel{y}{_x\triangle_{z}}$. If one vertex of the triangle is left blank, the net value of the expression is the value needed to fill in that blank. This has the niceness of displaying the trinary relationship between the three values. Also, the left-to-right flow agrees with the English way of verbalizing these expressions. It does seem to make inverse identities awkward:

$\log_x(x^y)=y$ becomes $\stackrel{}{_x\triangle_{\stackrel{y}{_x\triangle_{\phantom{z}}}}}=y$. (Or you could just say $\stackrel{}{_x\stackrel{y}{\triangle}_{\stackrel{y}{_x\triangle_{\phantom{z}}}}}$.)

$x^{\log_x(z)}=z$ becomes $\stackrel{\stackrel{}{_x\triangle_{z}}}{_x\triangle_{\phantom{z}}}=z$. (Or you could just say $\stackrel{\stackrel{}{_x\triangle_{z}}}{_x\triangle_{z}}$.)

$\sqrt[y]{x^y}=x$ becomes $\stackrel{y}{\triangle}_{\stackrel{y}{_x\triangle_{\phantom{z}}}}=x$. (Or you could just say $\stackrel{}{_x\stackrel{y}{\triangle}_{\stackrel{y}{_x\triangle_{\phantom{z}}}}}$ again.)

$(\sqrt[y]{z})^y=z$ becomes $_{\stackrel{y}{_\phantom{x}\triangle_{z}}}\hspace{-.25pc}\stackrel{y}{\triangle}=z$. (Or you could just say $_{\stackrel{y}{_\phantom{x}\triangle_{z}}}\hspace{-.25pc}\stackrel{y}{\triangle}_z$.)

(I am sure that there must be better ways to typeset these, but this is what I could come up with.)

Having $3$ variables, I was sure that there must be $3!$ identities, but at first I could only come up with these four. Then I noticed the similarities in structure that these four have: in each case, the larger $\triangle$ uses one vertex (say vertex A) for a simple variable. A second vertex (say vertex B) has a smaller $\triangle$ with the same simple variable in its vertex A. The smaller $\triangle$ leaves vertex B empty and makes use of vertex C.

With this construct, two configurations remain that provide two more identities:

$_{\stackrel{y}{_\phantom{x}\triangle_{z}}}\hspace{-.25pc}\stackrel{}{\triangle_z}=y$ states that $\log_{\sqrt[y]{z}}(z)=y$.

$\stackrel{\stackrel{}{_x\triangle_{z}}}{_\phantom{x}\triangle_{z}}=x$ states that $\sqrt[\log_x(z)]{z}=x$.

I was questioning the usefulness of this notation until it actually helped me write those last two identities. Here are some other identities:

\begin{align} \stackrel{a}{_x\triangle_{\phantom{z}}}\cdot\stackrel{b}{_x\triangle_{\phantom{z}}}&={}\stackrel{a+b}{_x\triangle_{\phantom{z}}}& \frac{\stackrel{a}{_x\triangle_{\phantom{z}}}}{\stackrel{b}{_x\triangle_{\phantom{z}}}}&={}\stackrel{a-b}{_x\triangle_{\phantom{z}}}& _{\stackrel{a}{_x\triangle_{\phantom{z}}}}\hspace{-.25pc}\stackrel{b}{\triangle} &={}\stackrel{ab}{_x\triangle_{\phantom{z}}}\\ \stackrel{}{_x\triangle_{ab}}&={}\stackrel{}{_x\triangle_{a}}+\stackrel{}{_x\triangle_{b}}& \stackrel{}{_x\triangle_{a/b}}&={}\stackrel{}{_x\triangle_{a}}-\stackrel{}{_x\triangle_{b}}&\stackrel{}{_x\triangle}_{\stackrel{b}{_a\triangle_{\phantom{z}}}}&=b\cdot\stackrel{}{_x\triangle}_{a} \\ \stackrel{-a}{_x\triangle_{\phantom{z}}}&=\frac{1}{\stackrel{a}{_x\triangle_{\phantom{z}}}}& \stackrel{1/y}{_x\triangle_{\phantom{z}}}&=\stackrel{y}{_\phantom{x}\triangle_{x}}& \stackrel{}{_x\triangle_{1/a}}&=-\mathord{\stackrel{}{_x\triangle_{a}}}\\ \stackrel{}{_a\triangle_{b}}\cdot\stackrel{}{_b\triangle_{c}}&=\stackrel{}{_a\triangle_{c}}& \stackrel{}{_a\triangle_{c}}&=\frac{\stackrel{}{_b\triangle_{c}}}{\stackrel{}{_b\triangle_{a}}} \end{align}

• You can go simpler. Consider " $b \stackrel{p}{\lrcorner} r$ ", with "$b$" the base, "$p$" the power, and "$r$" the result (for lack of a better word), with a fill-in-the-blank philosophy. Note that the symbol points to the components that "create" the result, making a visual connection (and breaking the 3-way symmetry). Interestingly, " $b \stackrel{p}{\lrcorner}$ " resembles " $b^p$ " (we can say the "$\lrcorner$" is "understood"); and " $\stackrel{p}{\lrcorner} r$ " is reminiscent of "$\sqrt[p]{r}$"; also, " $b \lrcorner r$ " has a (backwards, or tipped-over) "L", for "logarithm". :) – Blue Jul 3 '12 at 22:40
• Your idea got made into a YouTube video called the Triangle of Power – Mark May 4 '16 at 12:41
• Excellent video names this notation Triangle of Power by the illustrious 3Blue1Brown. – Bob Stein May 4 '16 at 12:43
• If it's the Triangle of Power, we need a Triangle of Courage and Triangle of Wisdom to go along with it, to complete the Triforce. – alex.jordan May 4 '16 at 17:59
• The triangle notation will make the functions look (at a glance) very similar and sort of squash their, shall we say, individual personalities, i.e. once you get to know log, you know log behaves in a certain way and you expect certain things from it as a function so that if someone says: Does "$\sum_{n=1}^{\infty} 1/\log n$ converge? You just say "no, log is too slow". You can't reason like that if you are stuck thinking "reciprocal of the number you raise the number e to in order to get n" – Thompson Jul 31 '16 at 13:59

Converting a comment to an answer (my third for this question!), by request. I think it might actually constitute my best suggestion.

Consider $$b\stackrel{p}{\lrcorner}r$$ with $b$ the base, $p$ the exponent, and $r$ the result (for lack of a better word (see below)), with a fill-in-the-blank philosophy: whatever's missing is what the symbol represents.

\begin{align} b\stackrel{p}{\lrcorner} &\quad:=\quad \text{the result from base b with exponent p}&\text{(aka "the p-th power of b")} \\ \stackrel{p}{\lrcorner}r &\quad:=\quad \text{the base giving result r from exponent p}&\text{(aka "the p-th root of r")} \\ b\lrcorner{r} &\quad:=\quad\text{the exponent yielding r with base b}&\text{(aka "the base-b logarithm of r")} \end{align}

Interestingly, "$b \stackrel{p}{\lrcorner}$" resembles "$b^p$"; we can say that the "$\lrcorner$" is "understood". Also, "$\stackrel{p}{\lrcorner} r$" is reminiscent of "$\sqrt[p]{r}$". One might even say that "$b \lrcorner r$" incorporates a backwards (or tipped-over) "L", for "Logarithm". :)

Note that the symbol points to the components that create the result (again, see below), and makes for a nice visual mnemonic: the flat part points to the base; the upward part points to the exponent to which the base is raised. This being so, I think I'd allow the "$\lrcorner$" symbol to be reversed, if someone had need: $$\stackrel{p}{\lrcorner} r \;\equiv\; r \stackrel{p}{\llcorner} \qquad\qquad b \lrcorner r \;\equiv\; r \llcorner b \qquad\qquad b\stackrel{p}{\lrcorner} \;\;\equiv\;\; \stackrel{p}{\llcorner}b$$

The re-orderability of $b$ and $r$ could come in handy, for instance, if one or the other involved a particularly-cmplicated expression. Anyway, the point is that the symbol --in either orientation-- makes clear what the roles of the components are.

(For optimal flexibility, we could make the symbol's "base" arm visually distinct from its "exponent" arm, say, with a double-bar in that dirction or something. (A cursory scan of the "Comprehensive LaTeX Symbol List" didn't reveal anything I liked.) Then you could orient the symbol and its attached components any way you pleased.)

Terminology. As @alex.jordan remarks in a comment to my comment to his answer, "[my] explanation is biased towards exponentiation over roots and logs". I don't disagree, especially with my use of the word "result" for component $r$. That said, I wrote "result" with the disclaimer "for lack of a better term" because ... well ... I lacked a better term. Almost two years later, I still do. Perhaps now is the time to confront the issue.

The Math Forum's Dr. Math makes the case that the result of an exponentiation is properly called a "power" ---think "the $3$rd power of $4$ is 64"--- and that we're playing fast and loose with terminology when we use "power" and "exponent" interchangeably. Fair enough. (Accordingly, I corrected my prose when converting it from my previous comment, and I'll make a conscious effort to be more careful in the future.) However, given that we do tend to use "power" and "exponent" interchangeably, I can't quite bring myself to call $r$ a "power" in conjunction with my notation.

But what, then?

In "$\sqrt[p]{r}=b$", component $r$ is the "radicand" $r$; in "$\log_b r = p$", it's the "argument". The latter is generic function-jargon with no specific meaning in the current context; the former, on the other hand, is hyper-specific, having been invented for its purpose. These terms offer us no guidance. I'll note that "sum" and "product" connote the result of an addition or a multiplication (sometimes both! See Jeff Miller's "Earliest Known Uses..." entry for "product"). Maybe we can obscure the distasteful bias of "result $r$" beneath some profound-sounding Latin derivative.

Any suggestions?

• I also had the same idea that qiaochu had - the symmetries of the symbol should reflect the symmetries of the operation. It needs to be a symbol with 3 "input" areas, that isn't symmetric under any rotation or reflection. I came up with this operator The advantage is that the new notation for nth root looks a lot like the old notation. Obviously logs are totally changed - but IMO logs need a symbol rather than a word anyway, and giving logs an easy, unique infix symbol is a massive win. – Rationalist Jul 28 '16 at 13:03
• This is the symbol I am suggesting: $$\llap{\surd}\backslash$$ So we have $$a \hspace{4 mm} \llap{\surd}\backslash \hspace{1 mm} b = log_{a}(b)$$ – Rationalist Jul 28 '16 at 13:38
• $$a \hspace{3mm} \stackrel{b}{\llap{\surd}\backslash}c$$ – Rationalist Jul 28 '16 at 13:46
• @Rationalist: Well, I did suggest making the arms of my symbol visually distinct. :) Over time, I've thought that, instead of a "double-bar" on the base arm, I'd prefer making the exponent arm more of a "harpoon". Something like this ... $$\huge{\_\kern{-1em}\upharpoonleft}$$ ... but without the little downward tail where horizontal meets vertical. Ditching the double-bar avoids having to lift one's "pen" for an extra stroke; moreover, the harpoon amplifies the vernacular of "raising" the base by the exponent. (BTW: It's interesting that this five-year old question still generates traffic!) – Blue Jul 28 '16 at 14:21
• nice. What drove me to $$a \hspace{3mm} \stackrel{b}{\llap{\surd}\backslash}c$$ is that $$\hspace{3mm} \stackrel{b}{\llap{\surd}\backslash}c = \sqrt[b]{c}$$ which I feel really eases the transition from the old notation to the new. It is indeed a pleasure to work with this notation. All those nasty log theorems come out so easily. – Rationalist Jul 28 '16 at 14:37

Just "thinking out loud" here ...

If we take the inline notation "$x$^$y$", and we emphasize the notion of "^" as raising to the power of $y$, then we might exaggerate the upward arrow, thusly:

$$x\stackrel{y}{\wedge} \;\; = z$$

In that case, roots amount to lowering from the power of $y$:

$$z\stackrel{y}{\vee} \;\; = x$$

The inverse nature of the operations then becomes clear, because "raising" and "lowering" cancel:

$$x\stackrel{y}{\wedge}\stackrel{y}{\vee} \;\; = x\stackrel{y}{\vee}\stackrel{y}{\wedge} \;\; =x$$

(Of course, they don't cancel so cleanly when $x$ is negative (or non-real).)

More generally, the rules of composition are pretty straightforward:

$$x\stackrel{a}{\wedge} \stackrel{b}{\wedge} \;\; = x \stackrel{ab}{\wedge} \hspace{0.5in} x\stackrel{a}{\vee}\stackrel{b}{\vee} \;\; =x\stackrel{ab}{\vee}$$ $$x\stackrel{a}{\wedge} \stackrel{b}{\vee} \;\; = x \stackrel{a/b}{\wedge} \;\; = x\stackrel{b/a}{\vee}$$

and we can observe properties such as the commutativity of "$\wedge$"s and "$\vee$"s (again with a suitable disclaimer for negative (or non-real) $x$).

Is this better than the standard notation? I think there's some visual appeal here, but I doubt the mathematical community is inclined to start including giant up-arrows beneath their exponents; nor are down-arrows likely to be adopted when it's easier to write reciprocated exponents. But perhaps there's something in this that might help ease students into the lore of powers and roots.

If nothing else, the "lowering" notation is reminiscent of the standard root notation $$\sqrt[y]z \hspace{0.5in} \leftrightarrow \hspace{0.5in} \stackrel{y}{\vee} \; \overline{z} \hspace{0.5in} \leftrightarrow \hspace{0.5in} z \stackrel{y}{\vee}$$

with the "$y$" positioned within a downward-pointing arrow, so perhaps this helps satisfy your need for a visual connection in the standard notation.

As for logarithms ... I got nothin' (yet!).

• Blue, I like this notation, but I really like the notation from your comment on alex.jordan's answer. You should add that as an answer as well if you have time. – Zaz Feb 21 '15 at 9:48
• @Josh: Now that you mention it, I kinda like that notation better, too. :) I'll convert it to an answer at some point. – Blue Feb 21 '15 at 10:28

What about \begin{align*} x^y &\rightarrow ~x^y \\ \sqrt[y]{x} &\rightarrow ~ ^yx \\ \log_y(x) &\rightarrow ~ _yx \end{align*} This has the same shape as the triangle notation. Pre-subsctipts and pre-superscripts are not used in other common notations. Although a pre-superscript could look like a regular superscript of the previous letter: $x^yz$ could mean $x^y\cdot z$ or $x\cdot\sqrt[y]{z}$ , so care with spacing would be needed in some contexts.

They are shorthands for the following

$$x^y = \exp(y \cdot \exp^{-1}(x)) = z$$

$$\sqrt[y]{z} = z^{\tfrac{1}{y}} = \exp(\tfrac{1}{y} \exp^{-1}(z)) = x$$

$$\log_x(z) = \frac{\exp^{-1}(z)}{\exp^{-1}(x)} = y$$

Although the first two are uniform the sqrt notation is used to avoid writing fractions. Other than that the reason the notations are different is because they have their own algebraic laws (although they do mirror each other somewhat, due to being inverses).

By the way, exponentiation was probably invented first for naturals then integers then fractions before generalized to real numbers. For that reason the notations carry some "history" which isn't always a good thing.

If you want to use 'one' symbol, you could do something like:

$x^y = z$

$x=z^{\frac{1}{y}}$

So that you are using fractions in both cases, without invoking the root notation. When it comes to the third equality, you are starting with $x^y = z$ and are trying to isolate $y$. The way to do that is to take log base x of both sides -- that's the function that allows you to leave $y$ by itself and solve it. If you want a way of doing that using fractions (as in the previous two cases), to my knowledge there is no such way. If you are looking for a 'simpler'/more fitting symbol for the function, you can change log for anything you would like.

If you like it "visually" see it this way: The equation $x^y=z$ defines a surface $S$ in $(x,y,z)$-space. Depending on the situation one may view $S$ as a graph over the $(x,y)$-plane, the $(y,z)$-plane or the $(z,x)$-plane. Since $S$ has no obvious symmetries this gives rise to three completely different functions $(x,y)\mapsto z=f(x,y)$, $(y,z)\mapsto x=g(y,z)$, $(z,x)\mapsto y=h(z,x)$. Now instead of $f$, $g$, $h$ these functions are usually denoted in the familiar way you regret, the same way we write $x\cdot y$ instead of $p(x,y)$ when we take the product of $x$ and $y$.

One idea is to use $\exp_ba$ to mean $a^b$, $\exp_{1/b}a$ to mean $a^{1/b}=\sqrt[b]{a}$, and either $\exp_b^{-1}a$ or $\text{invexp}_ba$ to mean $\log_ba$; the point is that while raising to a power (using a given number as the base) does not require a new operation to "undo" it, exponentiation (using a given number as the exponent) does, known as the inverse of the exponential, or more commonly the logarithm.

• I like this, because it's pretty much standard math notation. – Mechanical snail Nov 26 '12 at 2:35

Let's try this again ...

(This is offered as a separate answer from my first, because it proposes something different.)

First, a bit of a digression: There's a slight difference in "feel" with notation for products and fractions. The expression "$x \cdot y$" asks directly "What is the result of multiplying $x$ and $y$?", which amounts to a straightforward computation. On the other hand $z/y$ --that is, the "inverse with respect to multiplication by $y$"-- asks indirectly "What value, multiplied by $y$, yields result $z$?"

Of course, the fraction "$z/y$" admits a handy interpretation as a straightforward computation: "What is the result of dividing $z$ by $y$?" ... although, when you really look at it, the computation has subtle alternative flavors: "Dividing $z$ into quantity-$y$ pieces yields a piece of what resulting size?" and "Dividing $z$ into size-$y$ pieces yields what resulting quantity?" This ambiguity is the result of the convenient commutativity of products: Since "$x \cdot y$" and "$y \cdot x$" amount to the same thing, it doesn't matter which number corresponds to "size" and which to "quantity". Despite the ambiguity, we somehow survive.

Now, with powers and roots and logarithms, we have same difference in "feel" ... but since the "direct" computation ("this, to that power") lacks commutativity, the flavors of the "indirect" inverse operations aren't so subtle; moreover --and more importantly-- those operations lack an intuitive(!) computational interpretation akin to "dividing" for fractions. (We often represent fractions with pizza slices; what's the pizza-slice picture for a fifth-root? Of a log-base-7?)

The point of all this is that it may be helpful to devise a notation that amplifies the direct-vs-indirect dichotomy, to try and make clear when the numbers in the notation provide pieces of a computational result, and when they express a puzzle in terms of the a result and one of the computational pieces.

For example, I'll keep the power notation from my previous answer:

$$x \stackrel{y}{\wedge}$$

This represents a direct computation: "$x$ raised to power $y$". The left-to-right nature of the symbol is important, for the proposed inverse (with respect to $y$) would appear as

$$\stackrel{y}{\wedge}\;z$$

The interpretation here --again reading left-to-right-- is that "(an implicit something) raising to power $y$ yields result $z$". This is the $y$-th root of $z$.

$$y \underset{x}{\wedge}$$

... for the direct computation "$y$, raising base $x$", and then ...

$$\underset{x}{\wedge}\; z$$

... for the indirect puzzle: "(and implicit something) raising base $x$ yields result $z$". This is the logarithm-base-$x$ of $z$.

That is, $\stackrel{y}{\wedge}$ always represents "raising to power $y$", and $\underset{x}{\wedge}$ always represents "raising base $x$". When these symbols are placed to the right of an argument, the argument is a part of a direct computation; when the symbols are place on the left of an argument, that argument is the result of a direct computation.

Although the notation succeeds in distinguishing direct and indirect concepts, I'm not really satisfied with it. The fact that $x^y$ is expressed in two different ways --$x\stackrel{y}{\wedge}$ and $y\underset{x}{\wedge}$-- is strange; and the canceling inverses doesn't seem as clean as it could be.

We could agree that down-arrows are inverses of up-arrows and leave things on the right:

$$\begin{eqnarray*} x \stackrel{y}{\wedge} &\hspace{0.25in}\leftrightarrow\hspace{0.25in}& \text{x raised to power y} \\ z \stackrel{y}{\vee} &\hspace{0.25in}\leftrightarrow\hspace{0.25in}& \text{z resulting from raising to power y} \\ y \;\underset{x}{\wedge} &\hspace{0.25in}\leftrightarrow\hspace{0.25in}& \text{y raising base x} \\ z \;\underset{x}{\vee} &\hspace{0.25in}\leftrightarrow\hspace{0.25in}& \text{z resulting from raising base x} \end{eqnarray*}$$

This way, inverses cancel and commute (disclaimers apply) more cleanly, as in my first answer, though we still have distinct ways of expressing $x^y$. It's a little weird to use down-arrows in notation that gets read in terms of "raising", but perhaps all that's needed there is a better symbol.

• In my head, I'm beginning to read "resulting from raising" as "via" (which seems appropriate, given the "$\vee$"). That is, the $y$-th root of $z$ is "$z$ via power $y$ [with base to be determined]"; and the base-$x$ log of $z$ is "$z$ via base $x$ [with power to be determined]". So, maybe the down-arrows for inverses aren't so bad, after all. – Blue Apr 17 '11 at 21:23

The simplest solution would be to use $$\wedge$$ and $$\vee$$, this is easy, fast, and the font doesn't get tiny:

$$e^x = \exp(x) = e\wedge x\\ \log(x) = e\vee x$$

It would be right associative:

$$e^{e^x} = \exp(\exp(x)) = e\wedge e\wedge x = e\wedge(e\wedge x) \\\log(\log(x)) = e\vee e\vee x = e\vee(e\vee x)$$

The invserse would be

$$e^{\log(x)} = \log(e^x) = e\vee e\wedge x = e\wedge e\vee x = x$$

Squares and exponential towers would be easier to read, with larger font:

$$x^2 = x\wedge 2 \\2^{2^{2^{\cdot^{\cdot^{\cdot}}}}} = 2\wedge2\wedge2\wedge\cdots$$

Exponent rules:

$$e^x\times e^y =e^{x+y} = e\wedge x\times e\wedge y = e\wedge(x+y) \\(e^x)^y = e^{xy} = (e\wedge x)\wedge y = e\wedge (xy)$$

You could even omit the parenthesis and write $$e\wedge xy$$.

We also introduce the notation for inverses: $$\overline{x} = \frac{1}{x}$$, square roots are now:

$$\sqrt{x} = x\wedge\overline{2}$$

And thus $$\sqrt{x}^2 = (x\wedge\overline{2})\wedge 2 = x\wedge(\overline{2}2) = x\wedge 1 = x$$.

Some familiar formulas:

\begin{align}(1)&&&\int_1^x \overline{x}\,\mathrm{d}x = e\vee x \\ (2)&&&b\vee a = \frac{d\vee a}{d\vee b} \\ (3)&&&e\wedge ix = \cos(x) + i\sin(x) \\ (4)&&&e\wedge x = \sum_{k=0}^\infty \frac{x\wedge k}{k!} \end{align}

I have also considered this question. I have not heard of an alternative notation, but have wondered why logs use letters rather than position and symbols.

I personally have thought that radical notation makes visual sense in that it is reminiscent of the symbol for long division. As exponentiation is repeated multiplication in its most basic sense, likewise roots are a form of repeated division.

For logarithms, I think it would make sense to place the base as a subscript before the power, just as exponents are superscript to the right of the base. An extended L could be added (as an inverted division symbol) to help emphasize the fact that logarithms are a form of proportional division. E.g.: $_2 |\underline 8 = 3$ says how many times does 2 go into 8, proportionally?

• I'm pretty sure that the radical symbol originates from the letter $r$, so the reminiscence, if any, is coincidental. Also, I don't see how roots are a form of repeated division... I would sooner view logarithms a such! In any event, I don't see how introducing new notation to replace awell-known one is a good idea. – tomasz Nov 2 '12 at 21:13
• Logs are different because for about 700 years, all arithmetic was done with logarithms. Brevity required writing things like $4.5 \times 3.2 = \text{antilog}( \log 4.5 + \log 3.2)$. Logarithms are, in a sense, too important to be regulated to a submit to the rules other operators would suggest. – DanielV May 21 '17 at 11:51

I love Day Late Don's vee-wedge notation. It's easy to remember $\wedge$ stands for exponentiation, while inverting it is the inverse operation. I'd like to go even further with that, and just use it as an operator symbol. If $a \times b$ is just $a$ added to itself $b$ times, and $a^{b}$ is just $a$ multiplied by itself $b$ times, why does exponentiation even deserve the fancy superscript notation? In fact, we can extrapolate (wrong term?) an infinite set of operators, creating each simply by saying it is equal to the last one applied to the same number ($a$) $b$ times, e.g. $a \times a$ repeated $b$ times is $a \wedge b$, $a \wedge a$ repeated $b$ times is $a$ 㫟 $b$, or whatever notation you want to use there, etc. Sorry if this doesn't answer anything for you.

• It's worth noting that the definition you used only gives $a (n) b$, where $(n)$ means the nth operator, for natural numbers $b$. If you want to extend this to rationals or reals, a lot more effort has to be put in. – Eric Stucky Jul 1 '12 at 6:17

You can use an explicit predicate and some kind of placeholder like $$\cdot$$ to select arguments to hoist out of the expression. let's use the three-place predicate $$E$$ to represent an exponential fact. This notation is inspired by internally headed relative clauses in some languages such as Navajo, but it's essentially just a more compact special case of set-builder notation.

$$E(x, y, z) \stackrel{\text{def}}{\iff} x^y = z \tag{101}$$

If we want to write $$2^3$$ , we write it like so (102):

$$E(2, 3, \cdot) \;\;\;\text{evaluates to}\;\;\; 8 \tag{102}$$

If we want to write $$\ln(7)$$, we write it like so (103):

$$E(e, \cdot, 7) \;\;\;\text{evaluates to}\;\;\; \ln(7) \tag{103}$$

To express the a cube root of 14 (like the principal root), we write (104):

$$E(\cdot, 3, 14) \tag{104}$$

This notation also admits an immediate generalization to extract more than one thing, for instance:

$$E(\cdot, \cdot, 4) \tag{105}$$

I think the most sensible interpretation for (105) is that it expands into a set of ordered pairs $$(x, y)$$ such that $$x^y = 4$$ , but you can also make it return an arbitrary pair instead similar to Hilbert's $$\varepsilon$$ operator (called $$\tau$$ in Bourbaki), which is more consistent with the single-cdot behavior.

The notation is unambiguous as long as we always interpret it as applying to a single named predicate, so (106) is ill-formed, but (107) is not. I'm using implies bottom instead of $$\lnot$$ because we could reasonably choose to have $$\lnot$$ bind more tightly to an expression than our implicit set-builder notation, and I'm trying to illustrate a point about resolving ambiguity in the notation.

$$\text{BAD!}\;\;\;\;\; E(\cdot, \cdot, \cdot) \to \bot \tag{106}$$

$$F(\cdot, \cdot, \cdot) \;\;\;\text{where}\;\;\; F(a, b, c) \stackrel{\text{def}}{\iff} E(a, b, c) \to \bot \tag{107}$$

There's another problem, which is that not every predicate will be able to uniquely determine all its parameters if all but one is missing. In fact, (104) required a convention in order to make the expression single-valued and deterministic. I'm not sure how to resolve this in general.

exp _b y = x

Basically, you would erase the word "log", which is completely non-descriptive of the process for newbies, and replace it with the word "exp" to represent that you are finding the exponent of base b that "gives you" y. The dash represents subscript, so in this case it's a subscript variable b.

An example:

exp _2 8 = x

Thus, x = 3.

A newbie would read this from left to right as "exponent of base 2 that gives us 8", which is 3.

Edit: This doesn't, unfortunately, unify the three ideas under a more indicative notation, but I think it does clarify the meaning of the logarithmic process. One could, of course, alter the letters; perhaps "expn" would be a better mnemonic device, or even "exponent needed", the latter, of course, being used only when first learning about logarithms; it would be shortened to "expn" (or something) later.

• This is a horrible idea because $\exp$ is common notation for something else. – YiFan Apr 18 at 13:41

I have an alternative to log, because when I read that square is a prolonged r i think in a prolonged L like this:

$${\large \mathcal{L}} \hspace{-0.4ex} \underline{\ x \,}$$

It's faster and clear when you are writing by hand, with a unique line. I place the base under the L. I also use these notation for trigonometrics function whit another words.

The code I use in latex is:

{\large \mathcal{L}} \hspace{-0.4ex} \underline{\ x \\,}

or

\newcommand{\loga}[1] {\mathord{\, \large \mathcal{L} \hspace{-0.4ex} \underline{\ #1 \\,}}}

• Underlining is almost never ever used anywhere in math. Using it for a function instead of parentheses is especially not a good idea. (Are you going to draw two underlines of different lengths on top of each other if you want to take the log of the log of something?) – YiFan Apr 18 at 13:45

## protected by Qiaochu YuanJul 3 '12 at 20:43

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