Is there a general solution to $a^x + b^x =c^x$ for $x\in\mathbb{R}$? I found this question online: Solve for $x$, given $6^x + 4^x =9^x$. A straighforward but  unsatisfactory estimation can be obtained by graphing the two lines (on Desmos, say) and reading out the result of the intersection.  Is there a more beautiful solution to this equation that one can arrive at without a graphing calculator? And is there a general solution to $a^x + b^x =c^x$ for $x\in\mathbb{R}$ and $a, b, c \in\mathbb{N}$?
 A: Dividing by $9^x$ gives you the equation $$(2/3)^x + (4/9)^x = 1,$$ and letting $y = (2/3)^x$, this is the quadratic equation $$y^2 + y = 1,$$ with unique positive solution $y = \frac{\sqrt{5} - 1}{2}.$ Taking logarithms gives you $x$.
This solution would not be nearly as nice for general $a,b,c.$
A: EDIT: This answer misinterprets the question to be about existence and uniqueness of the solution, not necessarily existence of a nice closed-form expression. I was not sure what exactly the OP wanted, so I will keep the answer for what it's worth.
Assuming $a, b, c \in \mathbb{R}$ are greater than $0$, and either $a \gt c, b \gt c$ or $a \lt c, b \lt c$ then you can still show there is a unique solution.
Divide by $c^x$ to obtain $(a/c)^x + (b/c)^x = 1$. The LHS is either a monotonically increasing or decreasing function of $x$. In both cases, for large positive $x$ and large negative $x$ we have that the expression goes to infinity or converges to $0$. Thus by the Intermediate Value Theorem there exists an $x$ where the LHS equals $1$. Since the function is (strictly) monotonic, the solution is unique.
In your specific example, we have the latter situation ($a < c, b < c$) and the expression on the LHS is $(6/9)^x + (4/9)^x$. Both $(6/9)^x$ and $(4/9)^x$ are strictly decreasing functions, and for large $x$ they eventually become smaller than $1/2$, while for large negative $x$ they become larger than $1$ so their sum takes on values smaller than $1$ and larger than $1$, so it must take the value $1$ for some $x \in \mathbb{R}$. The solution is unique because the function is strictly monotonic, and therefore injective.
A: AFAIK there is no closed-form solution to $a^x + b^x = c^x$ in general.  
With $x = \dfrac{t}{\ln(c/a)}$ and $r = \dfrac{\ln(b/a)}{\ln(c/a)}$, you can write the equation as
$$ 1 + \exp(r t) = \exp(t)$$
or $$t = \ln(1 + \exp(r t))$$
Thus $t = x \ln(c/a)$ is a fixed point of the function
$$f(t) = \ln(1 + \exp(rt))$$
and you might write $x$ as an infinitely nested expression
$$ x = \frac{1}{\ln(c/a)} \ln\left(1 + \exp\left( r \ln \left(1 + \exp\left(r \ln ( \ldots )\right)\right)\right)\right) $$
Or you might solve in terms of an infinite series in powers of $r$, which should converge if $r$ is small:
$$ t = \ln  \left( 2 \right) +{\frac {\ln  \left( 2 \right) }{2}}r+{\frac {
\ln  \left( 2 \right)  \left( \ln  \left( 2 \right) +2 \right) }{8}}{r
}^{2}+ \left( {\frac {3\, \left( \ln  \left( 2 \right)  \right) ^{2}}{
16}}+{\frac {\ln  \left( 2 \right) }{8}} \right) {r}^{3}-{\frac {\ln 
 \left( 2 \right)  \left(  \left( \ln  \left( 2 \right)  \right) ^{3}-
6\, \left( \ln  \left( 2 \right)  \right) ^{2}-36\,\ln  \left( 2
 \right) -12 \right) }{192}}{r}^{4}-{\frac {5\,\ln  \left( 2 \right) 
 \left(  \left( \ln  \left( 2 \right)  \right) ^{3}-6\, \left( \ln 
 \left( 2 \right)  \right) ^{2}-12\,\ln  \left( 2 \right) -{\frac{12}{
5}} \right) }{384}}{r}^{5}
+ \ldots $$
