Alternative notation for exponents, logs and roots?
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:
$$\begin{align} \stackrel{a}{_x\triangle_{\phantom{z}}}\cdot\stackrel{b}{_x\triangle_{\phantom{z}}}&={}\stackrel{a+b}{_x\triangle_{\phantom{z}}}& \frac{\stackrel{a}{_x\triangle_{\phantom{z}}}}{\stackrel{b}{_x\triangle_{\phantom{z}}}}&={}\stackrel{a-b}{_x\triangle_{\phantom{z}}}& _{\stackrel{a}{_x\triangle_{\phantom{z}}}}\hspace{-.25pc}\stackrel{b}{\triangle} &={}\stackrel{ab}{_x\triangle_{\phantom{z}}}\\ \stackrel{}{_x\triangle_{ab}}&={}\stackrel{}{_x\triangle_{a}}+\stackrel{}{_x\triangle_{b}}& \stackrel{}{_x\triangle_{a/b}}&={}\stackrel{}{_x\triangle_{a}}-\stackrel{}{_x\triangle_{b}}&\stackrel{}{_x\triangle}_{\stackrel{b}{_a\triangle_{\phantom{z}}}}&=b\cdot\stackrel{}{_x\triangle}_{a} \\ \stackrel{-a}{_x\triangle_{\phantom{z}}}&=\frac{1}{\stackrel{a}{_x\triangle_{\phantom{z}}}}& \stackrel{1/y}{_x\triangle_{\phantom{z}}}&=\stackrel{y}{_\phantom{x}\triangle_{x}}& \stackrel{}{_x\triangle_{1/a}}&=-\mathord{\stackrel{}{_x\triangle_{a}}}\\ \stackrel{}{_a\triangle_{b}}\cdot\stackrel{}{_b\triangle_{c}}&=\stackrel{}{_a\triangle_{c}}& \stackrel{}{_a\triangle_{c}}&=\frac{\stackrel{}{_b\triangle_{c}}}{\stackrel{}{_b\triangle_{a}}}& \stackrel{\stackrel{-n}{_y\triangle_{\phantom{z}}}}{_x\triangle_{\phantom{z}}}&=\stackrel{\stackrel{n}{_y\triangle_{\phantom{z}}}}{_\phantom{x}\triangle_{x}}& \end{align}$$
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?
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!).