Solving $2^x - 3^x + 6^x =0$. Are there any known methods to solve
$$2^x - 3^x + 6^x = 0,$$ where $x$ is either in closed form, perhaps in terms of special functions, or to give inequalities on the answers, where $x\in\mathbb{C}$ ?

My thoughts: $x\in\mathbb{C}$ would presumably lead to many solutions. My first attempt was the following:

$$e^{x\log 2}-e^{x\log 3}+e^{x\log 6}=0.$$ Letting $x=a+ib$ then gives the two equations

$$2^a\cos(b\log 2)-3^a\cos(b\log 3)+6^a\cos(b\log 6)=0$$
and
$$2^a\sin(b\log 2)-3^a\sin(b\log 3)+6^a\sin(b\log 6)=0,$$
corresponding to the real and imaginary parts respectively. From then on it gets a bit messy. My feeling is there are no known methods except numerical ones?
Equations like this have infinitely many complex solutions.

Let $a = \log 2-\log 3$ and $b = \log 6 - \log 3 = \log 2$. Then let $f(z) = \exp(az) + \exp(bz)$. We want to solve $f(z)=1$

Since the two coefficients are real and have opposite signs, for $\Re(z) \to \infty$, $f(z) \sim \exp(bz)$ and for $\Re(z) \to - \infty$, $f(z) \sim \exp(az)$.

In fact, if you note $r_+$ the positive real solution to $\exp(bx) = \exp(ax) +1$ and $r_-$ the negative real solution to $\exp(ax) = \exp(bx) + 1$ (in fact, it is $-1$), all the solutions to $f(z) = 1$ must be in the region $r_- \le \Re(z) \le r_+$

Then the number of solutions with $|\Im(z)| &lt; B$ is the number of turns around $1$ made by $f(z)$ as you go along the rectangle whose vertices are $r_\pm \pm iB$
(more precisely, it's $\frac 1{2i\pi} \int f'/(f-1) dz$)

On the right side of the rectangle, $f$ is homotopic to $\exp(bz)$ without going through the disk, so the integral is $2ibB + O(1)$ 

On the left side, same thing except we go downwards, so we get $-2iaB + O(1)$.

Summing up everything, assuming the top and bottom integrals are also a $O(1)$, the number of $1$s with $|\Im(z)| &lt; B$ is $\frac{(b-a)B}\pi + O(1) = \frac {B\log 3} \pi + O(1)$ .



For a more detailed analysis,

let $H(x) = a$ for $x&lt;0$ and $H(x) = b$ for $x&gt;0$, and define

$I(y_1,y_2) = \int_{\Bbb R} \left(\frac {a\exp(a(z-iy_1)) + b\exp(b(z-iy_2))}  {\exp(a(z-iy_1)) + \exp(b(z-iy_2)) - 1} - H(z)\right) dz$.

Since the integrand converges to $0$ exponentially at $\pm \infty$, this integral is well-defined whenever the integrand doesn't have a pole,
which means whenever we don't find a solution to the shifted equation on the real line. Away form those solutions, $I$ is even holomorphic asa function of two complex variables.

Moreover, if $dt &gt; 0$ is real, then a change of variables gives
 $I(y_1+idt,y_2+idt) = \int_{\Bbb R} \left(\frac {a\exp(a(z-iy_1+dt)) + b\exp(b(z-iy_2+dt))}  {\exp(a(z-iy_1+dt)) + \exp(b(z-iy_2+dt)) - 1} - H(z) \right) dz
 \\ = \int_{\Bbb R} \left(\frac {a\exp(a(z-iy_1)) + b\exp(b(z-iy_2))} {\exp(a(z-iy_1)) + \exp(b(z-iy_2)) - 1} - H(z-dt)\right) dz
 \\ = I(y_1,y_2) + \int_{\Bbb R} (H(z) - H(z-dt)) dz
 \\ = I(y_1,y_2) + \int_0^{dt} (b-a)dz
 \\ = I(y_1,y_2) + (b-a)dt$

Since this is true for all positive real $dt$ and $I$ is holomorphic, 
it is also valid for any complex $dt$ (as long as we don't go over a solution),
replacing $dt$ with $-idt$ gives us $I(y_1+dt,y_2+dt) = I(y_1,y_2) + i(a-b)dt$.

Moreover, as a function of two real variables, $I$ is $2\pi/a$-periodic in the first and $2\pi/b$-periodic in the second,
so we only need a nice picture of $I$ on a fundamental rectangle. 
And to do this we need the picture of where the integral jumps (by $2i\pi$) because of a solution to the shifted equation.

Let $x \in \Bbb R$. A solution at $z=x$ happens when $1 = \exp(a(x-iy_1)) + \exp(b(x-iy_2))$.
With the triangle inequality and the law of cosines, this is only possible when $x \in [r_- ; r_+]$
with $2\exp(ax)\cos(ay_1) = 1 + \exp(2ax) - \exp(2bx), 2\exp(ax)\cos(by_2) = 1 + \exp(2bx) - \exp(2ax), \sin(ay_1)\sin(by_2) \le 0$.

Plotting those points give this graphics :



Finding a solution at $x+iy$ to the original problem is equivalent to reaching a discontinuity (so reaching one of the black curves) at $I(y,y)$. For example this shows that there are two solutions with imaginary part in $[0; 2\pi/b]$ because the diagonal crossed two curves so far.

Since $a/b$ is irrationnal, when we follow that diagonal using $I(y_1+dt,y_2+dy) = I(y_1,y_2) + i(a-b)dt$ and adding back $2i\pi$ whenever you cross the curve, the trajectory is dense in the corresponding torus, and therefore by continuity, $I$ is determined by this up to an additive constant.

In fact, now we can check that $I(y_1,y_2) = I(0,0) + iay_1 - iby_2 \mod (2i\pi)$. So $I$ was secretly a piecewise affine function ! (though as you input complex arguments, the black curve will change shape and you might have more complicated things happening, for example it might cross itself ... )

We also see that if we "approximate" the curves with straight lines, then the number of solutions we find with imaginary part between $0$ and $B$ would be exactly $\lfloor (b-a)B/\pi + k \rfloor$ for some constant $k$. For the actual equation, this is still $(b-a)B/\pi + O(1)$ and you can find the best lower and upper bound for the $O(1)$ by measuring precisely how far the curve gets away from the approximation line, and how far the origin is from the first line.

To find the distribution of the real part of the solutions, (so the expected number of roots with real part in some given interval in $[r_-,r_+]$), thanks to Weyl's equidistribution theorem you "only" need to highlight the points of the curves corresponding to that interval and see how much "shadow" they generate when you project them along the diagonal compared to the rest of the curve. This is a bunch of ugly calculus so I'll skip it.
The four true and only solutions...