$$
\begin{array}{rl}
\text{power law:} & y = x^{(\text{constant})}\\
\text{exponential:} & y = (\text{constant})^x
\end{array}
$$
That's the difference.
As for "looking the same", they're pretty different: Both are positive and go asymptotically to $0$, but with, for example $y=(1/2)^x$, the value of $y$ actually cuts in half every time $x$ increases by $1$, whereas, with $y = x^{-2}$, notice what happens as $x$ increases from $1\text{ million}$ to $1\text{ million}+1$. The amount by which $y$ gets multiplied is barely less than $1$, and if you put "billion" in place of "million", then it's even closer to $1$. With the exponential function, it always gets multiplied by $1/2$ no matter how big $x$ gets.
Also, notice that with the exponential probability distribution, you have the property of memorylessness.
I'm no mathematician, and there may be more concrete definitions, but this is how I think of a function as exponential "growth" and "decay."
Theory
If an exponential function is "skyrocketing" (for lack of better terminology) and heads towards $\pm\infty$, then it's "growing" (you can think of it as "absolute" growth, and disregard the sign).
If an exponential function is "slowing down" (again, for lack of better terminology), and heads towards something other that $\pm\infty$ (i.e. has some horizontal asymptote), then it is "decaying."
Concrete
Is $ f(x) = -e^x $ exponential growth, decay, or neither?
You can see that as $x$ increases, $f(x)$ will head to $-\infty$. So, I'd call this a growth.
Is $ g(x) = -e^{-x} $ exponential growth, decay, or neither?
For easier visualization of a corresponding graph, let us rewrite $g(x) = -\frac{1}{e^x}$. You can see that as $x$ increases, $g(x)$ will head to $0$. So, I'd call this a decay.
Best Answer
The "Square is a rectangle" relationship is an example where the square is a special case of a rectangle.
"Exponential decay" gets its name because the functions used to model it are of the form $f(x)=Ae^{kx} +C$ where $A>0$ and $k<0$. (Other $k$'s above $0$ yield an increasing function, not a decaying one.)
Similarly for "logarithmic decay," it gets its name since its modeled with functions of the form $g(x)=A\ln(x)+C$ where $A<0$.
These two families of functions do not overlap, so neither is a special case of the other. The giveaway is that the functions with $\ln(x)$ aren't even defined on half the real line, whereas the exponential ones are defined everywhere.