Solve $a^x + b^x = c$ for $x$. This is something I came across today (a specific case of this equation), only to realize I don't have the necessary tools to attack this problem.
This is not any kind of homework or research problem or anything like that. I'm genuinely interested how to approach solving this sort of problems. It may be surprisingly simple, but I just don't see it.
Any help would great appreciated. 
 A: Since your original question is to solve
$$6^x - 2^x = 32$$
and since there is no closed-form solution for the general equation, let's focus on the specific case. First you can easily find a solution $x=2$, but what about others?
Let $f(x) = 6^x - 2^x -32 = 2^x\cdot(3^x-1) - 32$. Now you can apply Rolle's theorem to show that the only solution is $x=2$:


*

*the function is stricly increasing (its derivative is $f'(x) = 6^x \log (6)-2^x \log (2) = 2^x\cdot(3^x \cdot \log 6 - \log 2) > 0 \quad \forall x \in \mathbb{R^+}$

*$f(0) = -32 < 0, f(3) > 0$

*since it's a composition of continuous and differentiable functions, it is continuous and differentiable. Thus the theorem tells us that the only solution is a member of the interval $[0;3]$ so $x=2$ is the only solution.

A: Just as  Yves Daoust commented, in the most general case, this equation does not show explicit solutions and numerical methods (such as Newton method) will be required.
This does not make (in principle) any problem if a reasonable estimate can be used for starting iterations. 
So, let us consider $$f(x)=a^x+b^x-c$$ the zero of which being looked for. This function can be very stiff and, so, it would be better to consider $$g(x)=\log(a^x+b^x)-\log(c)$$ Let us assume $a >b>1$ and $c>0$; so, the solution would be such that $$\frac{\log (c)-\log (2)}{\log (a)}< x<\frac{\log (c)-\log (2)}{\log (b)}$$ which is already good since we bracketed the solution. 
So, let us be lazy and choose the midpoint as $x_0$.
Let us try using $a=\pi$, $b=e$, $c=123456789$. If you look at the plot of function $g(x)$, you should notice that it is very close to  straight line (which is always good).
 Using the above bounds, we already know that $$15.6703<x<17.9383$$ So let us start iterating with Newton methods : it will provide as successive iterates $$x_0=16.8042746089317$$ $$x_1=16.1960353781344$$  $$x_2=16.1957749465687$$ $$x_3=16.1957749465186$$ which is the solution for fifteen significant figures.
A: Considering the equation $a^x+b^x=c$, where $a>b>0$:
(Not $6^x-2^x=32$ from the comments.)
The derivative of the LHS is
$$f'(x)=\ln(a)a^x+\ln(b)b^x,$$
which has a single root,
$$x'=\log_{a/b}\left(-\frac{\ln(b)}{\ln(a)}\right)$$ when $a>1>b$, corresponding to a minimum of $f$, and none otherwise.
If $x'$ exists, the equation has two solutions iff $f(x')<c$; if $x'$ does not exist, it has a single solution iff $c>0$.
An efficient procedure to bracket the solutions is to start from $x=x'$ and try arguments $x\pm s$ where the step increases by doubling, $s=2^{k}$. Very quickly, $f(x\pm 2^k)$ will exceed $c$.
In the monotonic case, the comparison $f(0)=2>c$ tells you the sign of the solution and the direction to step to.
From there, you can use one of the dichotomic or Regula Falsi methods which will guarantee convergence. Newton's iterations may be faster, but guarantees on convergence are much harder to establish (and in case you cross the $x'$ border, you are stuck).
