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