Weird sum of exponentials Is there a way to solve this: 
$$3^x + 4^x + 5^x = 6^x?
$$
I used brute force for this, but I dunno the solution. Any idea would be a great help. :)
 A: $x=3$. Dividing by $6^x$ you obtain a left-hand side that is monotonous decreasing and a constant right-hand side. The left-hand side goes to $0$ as $x\rightarrow \infty$ and goes to $\infty$ as $x\rightarrow -\infty$. Therefore there is exactly 1 root. You can easily solve it numerically by bisection.
A: You have been given the solution and explained why there is exactly one root.
For the fun of it, let us consider the more general case of
$$a^x + (a + 1)^x + (a + 2)^x = (a + 3)^x$$ where $a$ is a non negative number (integer or real). For that, and I would suggest you use the trick when plotting, take logarithms and consider the function $$f=\log((a+3)^x)-\log(a^x + (a + 1)^x + (a + 2)^x)$$ each piece being almost a straight line $$f=x \log(a+3)-\log(a^x + (a + 1)^x + (a + 2)^x)$$ We can bound the solution since $$3 a^x \lt a^x + (a + 1)^x + (a + 2)^x\lt 3(a+2)^x$$ $$\frac{\log (3)}{\log (a+3)-\log (a+2)} < x <\frac{\log (3)}{\log (a+3)-\log (a)}$$ So, we could use any numerical method inside well defined bounds. Better for sure would be Newton method that we could start at the mid point of the interval.
For illustration purposes, let us try using $a=15$. The successive iterates of  Newton method would be
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 12.62308376 \\
 1 & 10.29832716 \\
 2 & 10.35504436 \\
 3 & 10.35508032
\end{array}
\right)$$ which is the solution for ten significant figures.
Going deeper into this problem, an analysis (for the range $0 \leq a \leq 1000$) shows that the solution is very well approximated using $$x\approx 0.609384\, a+1.22586$$ Applied to your case $a=3$ (which, as Hagen von Eitzen commented, is probably the only one leading to an integer solution), this empirical correlation would give, as an estimate, $3.05401$ which is not too bad. For $a=15$, the estimate would be $10.3666$ to be compared to the exact solution given above.
Since the correlation was built up to $a=1000$, let us try with $a=10000$ to see if it still works. For this case, Newton iterates will be
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 6095.064000 \\
 1 & 6095.008961 \\
 2 & 6095.008961
\end{array}
\right)$$  which is the solution for ten significant figures.
Now, you can quiz your friends or even your teacher.
