How to find closed form or numerical solution to general power fit equation? What is $x$ in $a^x + b^x = c^x$ where $a,b,c$ are integers? What if $a,b,c$ are real numbers? What mathematics is required? Is numerical approximation required? Unfortunately, I cannot use Mathematica's built-in Solve[] for this.
 A: As Xander Henderson commented, numerical methods are required (in general).
Consider that you look for the zero of $$f(x)=c^x-(a^x+b^x)$$ Graphing would give you an estimate to start with.
Definitely better would be to consider $$g(x)=x \log(c)-\log(a^x+b^x)$$ the graph of which looking probably like a straight line.
Take for example $a=12$, $b=17$, $c=24$. Graphing $g(x)$ shows a solution between $1$ and $2$ (in fact, close to $1.5$). Using
$$g'(x)=\log (c)-\frac{a^x \log (a)+b^x \log (b)}{a^x+b^x}$$  Let us be very lazy and start iterating using $x_0=0$. You would get as iterates
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0 \\
 1 & 1.335559569 \\
 2 & 1.391504913 \\
 3 & 1.391598964
\end{array}
\right)$$ Quite fast, isn't it ?
Edit
In fact, we could make the problem more general with as many terms as wished in the rhs, that is to say solve 
$$f(x)=c^x-\sum_{i=1}^n a_i^x$$ For conveniency, consider that the numbers $a_i$ are increasing $(a_1 < a_2 < \cdots < a_n)$. The correspoding $g(x)$ function is now $$g(x)=x\log(c)-\log\left(\sum_{i=1}^n a_i^x \right)$$ which is bounded by
$x\log(c)-\log(n a_1^x)$ and $x\log(c)-\log(n a_n^x)$ and then the solution we look for is bounded bewteen
$$x_1=\frac{\log (n)}{\log (c)-\log (a_1)} \qquad \text{and}\qquad x_2=\frac{\log (n)}{\log (c)-\log (a_n)}$$ Computing $g(x_1)$ and $g(x_2)$ and computing the parameters of the straight line joining these two points, the estimate of the solution is then given by $$x_0=\frac{x_2\, g(x_1)-x_1\, g(x_2)}{g(x_1)-g(x_2)}$$ Applied to the test case, this gives an estimate  $x_0=1.39885$. Starting from this value, Newton method would converge in very few iterations.
