Where $x=y^d$ how do I express $d$? Given:
$$
x = y^d
$$
How do I express $d$?
I was tempted to think that $d$ is equal to the $y$ root of $x$ but that is false.
 A: If $y$ is positive that's the definition of logarithm. 
$\log_y x = d\iff y^d=x $.
And $d=\log_y x=\frac {\ln y}{\ln x} $.
If $y $ is negative there $y^d=\pm |y^d|$ depending on whether d is odd or even or has an odd or even numerator and an odd denominator. (The statement is impossible if $d$ has an even denominator or is irrational.)
So $d=\frac {\ln |x|}{\ln |y|} $.
If $y= 0$ then $d $ cannot be determined as all $d $ (except zero) will have $0=0^d $.
A: I will work in $\mathbb{K}$ where $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$ - I will also use the notation given in the OP's question; further, I will give a solution where $x$ is any real/complex number that isn't $0\in\mathbb{R}$ nor $0_{\mathbb{C}}=(0,0)\in\mathbb{C}$.
Simply put, if $\mathbb{K}=\mathbb{R}$, then the comments/answers provided already are clear explanations of how to represent $d$, as $d=\text{Log}_{y}(|x|)$, where $x\in\mathbb{R}\backslash\{0\}$ - the capital "L" is explained below.
It is when $\mathbb{K}=\mathbb{C}$ that things become more delicate. The overall goal here is to define $\log(x)$ as the inverse of the exponential function (remember we are working in $\mathbb{C}$, and the lower-case "l" signifies we are in this number system) -- I will build the definition of $\log(x)$ where the base is Euler's number $e$ (meaning $y=e$), and then go from there to establish what you are asking in $\mathbb{C}$. This way, you have both real-, and complex-, valued answers.
Anyways, we aim to define $\log(x)$ as the inverse of the exponential function; i.e., we wish to show:
(1) $~~$ $d=\log(x)$, $~~$ if $x=e^{d}$.
Now since $e^{d}$ is never zero, we can assume that $x\neq 0_{\mathbb{C}}$ (where $0_{\mathbb{C}}=0+0i$ with $i=(0,1)\in\mathbb{C}$). To find $\log(x)$ explicitly, we first write $x$ in polar form as $x=re^{i\theta}$, where $r\in(0,+\infty)$ and $\theta$ is the angle formed from the real-axis going counterclockwise to the vector $x=(u_{0},v_{0})\in\mathbb{C}$ in the complex plane; we also write $d$ in standard form as $d=u+iv$. The equation $x=e^{d}$ becomes:
(2) $~~$ $x=re^{i\theta}=e^{u+iv}=e^{u}e^{iv}$.
Taking the (complex) absolute value of both sides of (2), we now deduce that $r=e^{u}$, where $u$ is the ordinary (or real) logarithm of $r$, i.e., $u=\text{Log}(r)$ -- as it is mentioned above, we use the capital "L" to designate the natural logarithmic function of $real$ variables.
The equality of remaining factors in (2), namely $e^{i\theta}=e^{iv}$, identifies $v$ as the (multiple-valued) polar angle $\theta=\arg(x)$ (I will define the argument, multiple-valued, function, $\arg$, at the end): i.e., we have $v=\arg(x)=\theta$.
Thus $d=\log(x)$ is also a multiple-valued function (like the $arg$ function). The explicit definition is as followed:

Definition: If $x\in\mathbb{C}\backslash\{0_{\mathbb{C}}\}$ (i.e., $x\neq 0_{\mathbb{C}}$), then we define $\log(x)$ to be the set of infinitely-many values $\log(x)=\text{Log}(|x|)+i\arg(x)=\text{Log}(|x|)+i\big(\text{Arg}(x)+2k\pi\big)$, where $k\in\mathbb{Z}$.

In summary, in your question, if $y=e$ (or $y$ is Euler's number), then $d=\log(x)$. In this case, should $x\in\mathbb{R}\backslash\{0\}$, then we have that $\arg(x)=0\in\mathbb{R}$ (where we can assume $k=0\in\mathbb{Z}$ for convenience), and $d=\text{Log}(|x|)$. The argument function, $\arg$, stems from $x=(u_{0},v_{0})\in\mathbb{C}$ transformed into polar coordinates $(r,\theta)$ where $r=\sqrt{u_{0}^{2}+v_{0}^{2}}=|x|$ and $u_{0}=r\cos(\theta)$ with $v_{0}=r\sin(\theta)$. We can identify $\theta$ up to integer multiples of $2\pi$, so that the argument function is defined as a function taking infinitely-many values (i.e., a multiple-valued function) such that $\arg(x)=\{\theta+2k\pi:k\in\mathbb{Z}\}$. Observe that the $\arg$ function does contain discontinuities; for example, $\theta$ is taken in an interval of length $2\pi$, so, in particular, if our interval is $[0,2\pi)$ we have a jump of $2\pi$ when $\theta$ crosses over from $[0,2\pi)$ into $[2\pi, 4\pi)$, and so on. For convenience (and most used in application), we define $\text{Arg}(x)=\theta$, whenever $\theta\in[-\pi,\pi)$; note that these chosen intervals that are angle, $\theta$, is measured within are called $branches$, where the jump discontinuities are called $branch$-$cuts$, the interval $[-\pi,\pi)$ is called the $principle~branch$, and, lastly, the (single-valued) function $\text{Arg}$ is called the $principal~value~of~the~argument$.
Lastly, and with details omitted, if $y\neq e$, then we now also have the following definition:

Definition: If $d$ is any complex constant and $y\in\mathbb{C}\backslash\{0_{\mathbb{C}}\}$, then we define $y^{d}:=e^{d\log(y)}$.

We can now use everything above, to which everything now applies, and therefore if $y\in\mathbb{R}\backslash\{0\}$ we have what I referenced above -- namely that if $x=y^{d}$, then $d=\text{Log}_{y}(|x|)$.
A: From the basic properties of log,
$x = y^d$
implies
$\log(x) = d\log(y)$.
Therefore
$d 
=\dfrac{\log(x)}{\log(y)}
$.
The log can be taken to any base
(2, 10, $e$, ...)
because the division
cancals that out.
A: You know that we have the definition of logarithm as:
$$b=a^c \Rightarrow \log_a b=c$$
Furthermore, for an arbitrary based $\log$,
$$b=a^c \Rightarrow \log b=c\log a$$
As $\log a^b=b\log a$
Thus,
$$x=y^d\Rightarrow\log x=d\log y\Rightarrow d=\frac{\log x}{\log y}$$
This is for an arbitrary based log, if you like, it could also be expressed as:
$$d=\log_y x$$
