If we have

$$ x^y = z $$

then we know that

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


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

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.

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    $\begingroup$ 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. $\endgroup$
    – friedo
    Commented Mar 31, 2011 at 2:09
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    $\begingroup$ 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. $\endgroup$ Commented Jul 3, 2012 at 20:42
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    $\begingroup$ 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. $\endgroup$ Commented Jul 19, 2016 at 5:01
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    $\begingroup$ @LubošMotl I'm not sure that the current notation helps that, either. You can throw around $\exp x$ and $\ln x$ all you want, but until you explain derivatives of exponentials and the convergence of exponential growth, it's not going to make much sense anyway. Picture a 5th grader learning about exponents and radicals, or an eighth grader learning about logarithms. How would you explain to them why $e$ is a much better base than $10$? $\endgroup$
    – DonielF
    Commented Sep 14, 2018 at 18:57
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    $\begingroup$ But even if those things were only explained at the high school, they're still true. If you teach the kids to use a conceptually misguided notation, it will prevent them from understanding these things at the high school - which is still a serious enough problem. What's actually going on is that some of the folks don't understand $e$ and why it's more natural even as adults, and these people would like to determine education or mathematical notation. That's a path to eliminate mathematically literate kids from schools and from the future of nations. $\endgroup$ Commented Sep 16, 2018 at 5:56

17 Answers 17


Always assuming $x>0$ and $z>0$, how about: $$\begin{align} x^y &={} \stackrel{y}{_x\triangle_{\phantom{z}}}&&\text{$x$ to the $y$}\\ \sqrt[y]{z} &={} \stackrel{y}{_\phantom{x}\triangle_{z}}&&\text{$y$th root of $z$}\\ \log_x(z)&={} \stackrel{}{_x\triangle_{z}}&&\text{log base $x$ of $z$}\\ \end{align}$$ 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 $_{\stackrel{y}{_\phantom{x}\triangle_{z}}}\hspace{-.25pc}\stackrel{y}{\triangle}_z$.)

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:

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

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    $\begingroup$ @QiaochuYuan Would a scalene triangle be better then? We still have a trinary relationship to express, so a triangle seems like the simplest symbol. $\endgroup$
    – 2'5 9'2
    Commented Jul 3, 2012 at 20:46
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    $\begingroup$ 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". :) $\endgroup$
    – Blue
    Commented Jul 3, 2012 at 22:40
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    $\begingroup$ Your idea got made into a YouTube video called the Triangle of Power $\endgroup$
    – Mark
    Commented May 4, 2016 at 12:41
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    $\begingroup$ Excellent video names this notation Triangle of Power by the illustrious 3Blue1Brown. $\endgroup$
    – Bob Stein
    Commented May 4, 2016 at 12:43
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    $\begingroup$ 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" $\endgroup$
    – Thompson
    Commented Jul 31, 2016 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?

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jul 28, 2016 at 13:03
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    $\begingroup$ This is the symbol I am suggesting: $$\llap{\surd}\backslash$$ So we have $$a \hspace{4 mm} \llap{\surd}\backslash \hspace{1 mm} b = log_{a}(b)$$ $\endgroup$ Commented Jul 28, 2016 at 13:38
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    $\begingroup$ $$a \hspace{3mm} \stackrel{b}{\llap{\surd}\backslash}c$$ $\endgroup$ Commented Jul 28, 2016 at 13:46
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    $\begingroup$ @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!) $\endgroup$
    – Blue
    Commented Jul 28, 2016 at 14:21
  • $\begingroup$ 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. $\endgroup$ Commented Jul 28, 2016 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!).

  • $\begingroup$ 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. $\endgroup$
    – Zaz
    Commented Feb 21, 2015 at 9:48
  • $\begingroup$ @Josh: Now that you mention it, I kinda like that notation better, too. :) I'll convert it to an answer at some point. $\endgroup$
    – Blue
    Commented Feb 21, 2015 at 10:28
  • $\begingroup$ If you write down the n-th root of a as a^(1/3) we have basically the same behaviour as your notation, but more intuitive and with no additional symbols... $\endgroup$
    – Nearoo
    Commented Jan 25, 2020 at 8:52
  • $\begingroup$ @‍Blue's comment on @‍alex.jordan's answer, referenced by @Zaz; and the requested answer. $\endgroup$
    – LSpice
    Commented May 19, 2021 at 19:06

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 inverse 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}


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.

  • $\begingroup$ Can you please explain exp() and exp-1()? $\endgroup$ Commented May 3, 2016 at 13:54
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    $\begingroup$ Hysterical reasons are the worst. $\endgroup$ Commented May 4, 2016 at 11:37
  • $\begingroup$ Isn't $exp^{-1}(x)$ the same as $ln(x)$ in this case? It seems like a cop-out to use natural log in the definition of log. Ditto with using $exp$ in the definition of exponentiation. $\endgroup$
    – fpf3
    Commented Jun 21, 2023 at 14:09

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.

  • $\begingroup$ I like this, because it's pretty much standard math notation. $\endgroup$ Commented Nov 26, 2012 at 2:35


The question asked is "Has anyone ever considered alternative notation?" I think that it almost certain that many people have, but it is equally certain that no such notation has caught on. Most of the other answers here discuss proposed notations, none of which seem to have any traction (or even use) in the wider mathematical community.

As such, I am presenting this answer as a frame challenge, which is meant to address what I perceive as the underlying problem, as well as some of the misconceptions which have prompted this question.

Misconceptions in the Question

In the question, it is asserted that if $x^y = z$, then "we know that" $$ x = \sqrt[y]{z} \qquad\text{and}\qquad y = \log_x(z). $$ This is incorrect.

  • If we assume that $x$ is a real variable, that $n$ is a given natural number, and that $a$ is a real number, then the equation $$ x^n = a \tag{1}$$ has $n$ complex solutions. If $n$ is odd, one of those solutions will be real; if $n$ is even and $a > 0$, then two of those solutions will be real. The notation $\sqrt[n]{a}$, depending on context, either denotes the real solution to (1) (if $n$ is odd), the nonnegative real solution to (1) (if $n$ is even and $a > 0$).

    It is not obvious what $\sqrt[n]{a}$ should mean if $n$ is not a positive integer, though there are reasonable ways of defining this notation in terms of (1). For example, if $n$ is a natural number, we could define $$ \sqrt[-n]{a} = a^{-n} = \frac{1}{a^n},$$ but we don't typically do that, and instead rely on exponential notation alone.

    More generally, if $x$ is a complex variable, and $a$ and $n$ are complex constants, then $\sqrt[n]{a}$ typically denotes the principal $n$-th root of $a$, which can be defined in couple of slightly different ways, but generally means something like $r \mathrm{e}^{i/n}$. However, in this setting, there is enough ambiguity that one would usually choose to use more explicit notation in terms of the complex logarithm / exponential.

  • If we assume that $y$ is a real variable and that both $a$ and $b$ are positive real constants, then the equation $$ b^y = a \tag{2}$$ has a unique solution: $$y = \log_b(a) = \frac{\log(a)}{\log(b)}, $$ where $\log$ denotes the natural logarithm (or the common logarithm, or any logarithm with a fixed base—it doesn't really matter). On the other hand, as soon as we allow either $a$ or $b$ to be something other than a positive real number (say, a negative real number, or a complex number), things immediately become much more complicated, and (2) has many solutions (or none at all, depending on context).

In either case, we can make sense of the assertions stated in the original question if we first restrict the sets of numbers we are willing to consider. The suite of conclusions stated hold if we assume that $x$ and $y$ are positive real numbers and that $n$ is a natural number.


It is worth noting that notations adopted here come from a few very different historical antecedents. It is only in relative recent mathematical history that the link between exponential, logarithmic, and radical functions was well understood and articulated. The notation reflects this.

While I am not an expert in this area, my understanding is that something like the following is true:

  • Radical notation stems from classical Greek thinking (this is not to say that the Greeks used this notation; only that it reflects their way of thinking about problems). To the Greek way of thinking, a number represented a physical quantity—a length, or an area, or a volume. A number is inherently represented by a length, multiplication of two lengths gives an area, and so on. In this mode of thinking, it is very reasonable to ask "If the area of a square is $a$, what is the length of each side of that square?"

    In this framework, $a$ is a positive number, and the exponent ($2$), is a natural number. The square root of $a$ is then that side length (a positive number). Similarly, the cube root of $a$ is the length of the side of a cube which has volume $a$ (again, both the cube root and $a$ are positive numbers).

    The notation $\sqrt[n]{a}$ reflects this history—unless one has explicitly noted otherwise, $n$ is a natural number, and $a$ is a positive real number. We certainly can extend the definition of the radical notation, but my impression is that this is rarely done.

  • While this is the bit that I am most uncertain about, my understanding is that a more broadly defined real exponential function, i.e. $x \mapsto \exp_a(x)$, comes about with the rise of calculus in the mid-17th century. In this context, we assume that $a > 0$ and that $x$ is a real variable, which allows us to talk about rates of changes which are proportional to the underlying variable (e.g. the rate at which a colony of bacteria grow is proportional to the size of that colony: $$ \frac{\mathrm{d}P}{\mathrm{d}t} = kP, $$ where $k$ is some intrinsic growth rate).

    In this setting, $\exp(x)$ (the natural exponential function) is the unique solution to the initial value problem $$ \frac{\mathrm{d}y}{\mathrm{d}x} = y, \qquad y(0) = 1. $$ It can be observed that $\exp(x)$ has a lot of the same properties as $\mathrm{e}^x$ (where the former notation indicates the solution to an IVP, and the latter notation indicates "repeated multiplication"), but showing that these are the same requires a little bit of work.

    As such, it is probably healthy to use different notation for these two notions.

  • Logarithms were first developed in the 16th century by John Napier. In modern language and notation, Napier observed that there is a natural isomorphism between the multiplicative group of positive real numbers, and the additive group of all real numbers. As such, if you wanted to multiply two real numbers, it might save you some time to look up the logarithms of those two numbers in a table, add the results, and then take an antilogarithm (find the number in your table of logarithms whose logarithm is your sum).

    Addition and two or three table lookups are relatively quick when compared to multiplying two very large or very precise numbers, so books of logarithms proved to be quite useful. The fundamental idea is that a logarithm is a function $f$ which satisfies the functional equation $$ f(x+y) = f(x)f(y). $$ Napier's tables of logarithms took $f$ to be the common logarithm (that is, the logarithm with base 10), but it turns out that any log will do.

    Again, it is possible to show that the exponential function with base $a$ and the logarithmic function with base $a$ are inverse to each other, but this is a later historical development.


Thus far, I have claimed that the notations $a^x$, $\sqrt[n]{a}$, $\exp_a(x)$, and $\log_a(x)$ have distinct historical motivation, and were originally understood as representing very distinct notions. However, this does not necessarily mean that we should continue to regard them as distinct, nor that would should not adopt common notation.

Indeed, this brings me to the crux of what I believe this question is about, and to the nut of my answer as a frame challenge. The underlying question is not "Has anyone considered alternative notation?", but rather "Why don't we use and/or teach alternative notation?" and, perhaps, "Should we use alternative notation?"

An entirely correct, but also useless, answer is that we don't use alternative notation because we don't. Expanding on this a little, mathematical notation is a kind of language which we use to transmit ideas. Like all language, mathematical notation evolves over time, and is the product of human interaction. We don't use alternative notation because (a) the notation we have is sufficiently well understood by "fluent" or "native" speakers of mathematics, and (b) because the community of people who practice mathematics have never felt a need for such notation—it simply hasn't proved useful enough to dispense with the old notation.

In other words, the language of mathematical notation has not evolved a new set of notation for these relationships because the native speakers of that language have not felt a need for it.

Which brings us to the neophyte speakers—the students who might find a "unified" notation "easier to learn".

In principle, an instructor could introduce a new notation and teach that to students. Indeed, I have sometimes been tempted to dispense with $\pi$ and instead use the notation $\tau$ ($=2\pi$) when teaching trigonometry.

However, I think that such an action would ultimately do an incredible disservice to students. The goal of mathematics instruction is (or, at least, should be) to teach students to "do" mathematics as it is currently "done" by professionals in the community. Part of this requires that we teach students to use the language and notation of actual working mathematicians.

As such, students need to be familiar (and even comfortable) with exponential notation, logarithmic notation, function notation, radical notation, and so on. They should be taught the subtle distinctions between these different notations, and should understand when and why one notation might be preferable over another.


To completely unify the notation, we can start by writing $$ z = x^y = \exp(y \operatorname{Log}(x) + i2k\pi) \qquad\text{or}\qquad \exp(\operatorname{Log}(z) + i2k\pi) = \exp(y \operatorname{Log}(x)), $$ where we assume that $x,y,z\in\mathbb{C}$, $\operatorname{Log}$ is the principal branch of the complex logarithm, $\exp$ is the complex exponential function, and $k$ is any integer.

In this notation, we get something like[1] $$ x = \exp\left( \frac{\operatorname{Log}(z) + i2k\pi}{y}\right), $$ and $$ y = \frac{\operatorname{Log}(z) + i2k\pi}{\operatorname{Log}(x)}. $$

In other words, there is existing notation which already unifies the various notation of exponentiation, roots, and logarithms. It isn't necessarily "pretty", and it isn't appropriate for elementary students, but it already exists.

[1] I will note that I have been a little sloppy in solving for $x$ and $y$ under the assumption that $x$, $y$, and $z$ are complex. What I have written should be fine if $y$ and $z$ are real, but I was not too careful about chasing complex exponents around.


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

$x^y = z$


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.


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.

  • $\begingroup$ This could get confusing if you expressed something like (x^y)z $\endgroup$
    – 0x777C
    Commented Dec 2, 2020 at 13:08
  • $\begingroup$ @0x777C, re, parentheses are always available for disambiguation; just as you write (x^y)z to avoid confusion with x^(yz), you could also write $(x^y)z$ (currently unnecessary, but would become ambiguous with this proposal) to avoid confusion with $x({^yz})$. $\endgroup$
    – LSpice
    Commented May 19, 2021 at 19:04

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


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 $z$.

For exponentiation and logarithms, we could start with ...

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

... 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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Blue
    Commented Apr 17, 2011 at 21:23
  • $\begingroup$ Your two approaches to division have names in the math-education jargon: they are often called partitive and subtractive, where the partitive approach to $z/y$ asks "what is the size of each of $y$ equal pieces of $z$?" and the subtractive approach to the same problem asks "how many pieces of size $z$ can be made from a total of $y$?" $\endgroup$
    – LSpice
    Commented May 19, 2021 at 19:12

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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – tomasz
    Commented Nov 2, 2012 at 21:13
  • $\begingroup$ 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. $\endgroup$
    – DanielV
    Commented May 21, 2017 at 11:51

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.


"Minimal Traction Proposition"


  1. no "new" symbols

  2. minimal change in the writing conventions

Lets say we have some base_x^exponent_y=result_r, then:

$x^y = x^y$ . . . . . . . . no change here

$\sqrt[y]r = r^{\frac xy}$ . . . . . . . no change here either, I really prefer writing roots like this and thinking of "root of something" being a number with an exponent on (0,1) interval (using √ otherwise does not make much sense anyway). You will end up using parenthesis more often but I prefer that as well(a programer's degeneration)

$log_x(r) = x^?r$ . . . . just adding a question mark will imply that we are asking for the exponent, again the parenthesis might be used more often.

Anyway, 3Blue1Brown video took me here but I was thinking about the subject some time ago.

Personally I have two issues with "power triangle" approach:

major issue: although the triangle is visually pleasing, it is also somehow visually deceiving in the way, that while the triangle is equilateral, it kind of seems like the operations would be somehow proportionate which they are not. eg: 2^27 = "some huge number" so if there should be a triangle, and the angles should somehow correspond to this "disproportion". Given that there are always 180º in the triangle, the case of 2^27=r, the "triangle" will essentially become a line since the "result angle" will consume 99.9...% of the 180º, the same for 2^(1/9!) but the "line" shall be in other direction(now how do you solve that).

minor issue: vertical spacing is getting more spread out


That’s my proposal for a "unified notation". Differs from Alex.Jordan's in two respects:

  1. "It needs to be a symbol with 3 'input' areas, that isn't symmetric under any rotation or reflection". Rationalist
  2. It’s closer to standard notation.


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.

  • 4
    $\begingroup$ 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. $\endgroup$ Commented Jul 1, 2012 at 6:17

Ok, I've got one: notate addition with an underscore $$ 2 + 3 = 2 \text{ _ } 3 = 5 $$ and multiplication with an "upperscore" $$ 2 * 3 = 2 \thinspace \overline{\phantom{e\thinspace}} \thinspace 3 = 6 $$ Then $e^{2+3} = e^2*e^3$ becomes $$ e^{2 \text{ _ } 3} = e^{2} {_{\phantom{b}}^{\_}} e^{3} $$ the lhs is addition in the exponent, the rhs is multiplication in the base, the bar never has to change positions, which is nice and symmetrical. If we denote logarithms with subscripts $ln(x) = e_x$ then we get the same effect $$ e_{2\bar{}3} = e_2 \text{-} e_3 $$ The lhs is multiplication in the "subponent" (the argument of the $ln$), the rhs is addition in the base. I apologize for the rhs also looking like subtraction but standard Tex positioning algorithms don't put subscripts as low as they place superscripts high, which I won't bother combating right now, by hand though theres little issue.

One can imagine letting space be the operator and using (under/over)bars to indicate what the space does so that $\overline{2 \space 3 \space 4} = 24 $ and $\underline{2 \space 3 \space 4} = 9 $ and $2 x^2 + 3 x + 4 = \underline{\overline{2 \space x \space x} \space \overline{3 \space x} \space 4}$ (so called sum of products form). I don't advocate this obviously, its just a little offshoot idea.

If you're willing to stop using the usual fraction notation and you think superscripts and subscripts are awkward then maybe do this $$ \frac{2 \space 3}{e} = \frac{2 \space 3}{e \space e} $$ and this $$ \frac{e}{2 \space 3} = \frac{e \space e}{2 \space 3} $$ Though then theres no good way of telling what the base is... An interesting but probably bad idea.


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