When does $a^x = k\cdot x^b$ have a unique "nice" solution? I'm looking for an example of a transcendental equation with a unique "nice" solution that can be identified as correct by inspection.  My first thought was
$$3^x = x^3$$
but this equation has two solutions, one of which is trivial and the other of which requires $W$-functions.
However, it seems to me that if I introduce a constant factor on one side of the equation, I ought to be able to get the two function $f(x)=3^x$ and $g(x)=kx^3$ to be tangent to each other at a single point of intersection.  More broadly, it seems like if $a$ and $b$ are any positive numbers (and $b$ is odd) there should be some constant $k$ such that $a^x = kx^b$ has a unique positive solution.
I doubt there's a simple, closed form solution for this general problem that doesn't involve $W$ functions, but I don't really care about the general case; I just want one clear example.  So the question:
Can you provide a specific example of an equation of the form $x^a = k\cdot b^x$ for which $a, b, k$ are reasonably "nice", and the equation has a unique solution that's also reasonably "nice"?  Alternatively, can you confirm that no such example exists?
 A: Following your suggestion of tangency, we need
$$x_0^a=kb^{x_0}\ ,\qquad ax_0^{a-1}=kb^{x_0}\ln b$$
which implies
$$a=x_0\ln b\ .$$
Because of the logarithm, it seems unlikely that there are any "nice" possibilities other than $b=e$.  This leads to various possibilities such as
$$x^2=4e^{x-2}$$
which has a unique positive solution but also has a negative solution, and
$$x^3=27e^{x-3}$$
which perhaps will do what you want.
A: Telling you what you already know: $a^x$ "starts" at a value of $1$ (when $x=0$), whereas $k x^b$ starts at $0$. Since exponentials out-pace powers, if $k x^b$ were to cross $a^x$, from below to above, then it would have to cross back again later. Therefore, we must keep the functions from crossing at all: where they meet, they must be tangent. (Also, to keep the power below the exponential for $x<0$, we should require $b$ to be a odd integer, and $k$ to be some positive value.)
Okay, then ... There must be a place, say, $x=c > 0$, where the curves' values match and their slopes match, so we consider the relation and its derivative:
$$a^c = k c^b \qquad\text{and}\qquad a^c\log a = b k c^{b-1} \tag{1}$$
These imply
$$k c^b \log a = b k c^{b-1} \quad\to\quad c\log a = b \tag{2}$$
where $c$ and $a$ are such that $b$ is that odd integer we need. Therefore,
$$a^c = k c^{c\log a}\quad\to\quad k = \left(\frac{a}{c^{\log a}}\right)^c \tag{3}$$
giving the desired relation this form:
$$a^x = \left(\frac{a}{c^{\log a}}\right)^c x^{c\log a} \tag{4}$$
The solution, $x=c$, is as nice as you want it to be, although $a$ could be a little ugly to get that integer in $(2)$. Optimal niceness seems to be achieved by taking, $a=e$, so that $\log a = 1$, and we have

$$e^x = \left(\frac{e}{c}\right)^c\cdot x^c \tag{$\star$}$$

where we restrict $c$ itself to a nice odd integer. (Note that $c=1$ gives the nicest of all possible worlds.)
Here's a plot with $c=3$:

