Solving an equation in $x$, in which $x$ occurs as exponent four times Find the number of solutions to the equation
$$2011^x+2012^x+2013^x-2014^x=0$$
The answer seems to be zero, but I have no idea why. Please avoid considering complex solutions and other scary things.
 A: The answer can't be zero solutions.
Let $f(x)=2011^x+2012^x+2013^x-2014^x$
Obviously $f(x)$ is continuous (sum of continuous functions), and $f(0)=2$. But what about the behavior as $x\to\infty$?
A: By Laguerre's extension of Descartes' rule of sign,

The number of real roots of Dirichlet polynomials of the form
  $$\alpha_1 \beta_1^x + \alpha_2 \beta_2^x + \cdots + \alpha_n \beta_n^x
\quad\text{ subject to }\quad
\begin{cases}\alpha_1, \ldots, \alpha_n \ne 0\\\beta_1 > \beta_2 > \cdots \beta_n > 0\end{cases}
$$
  is no more than number of sign changes in the finite sequence $(\alpha_1, \ldots, \alpha_n)$.
In addition, counting multiplicity, the difference between the number of real roots and number of sign changes is an even number.

If we apply this to the Dirichlet polynomial
$$2014^x - 2013^x - 2012^x - 2011^x,$$
the number of sign changes is $1$ and hence the number of real roots is $1$.
For more info, you can either look up the original paper (in French)


*

*E Laguerre, Sur la théorie des équations numériques, J. Math. Pures et Appl. 9 (1883)


or a modern introduction of same subject


*

*G.J.O Jameson, Counting zeros of generalized polynomials: Descartes' rule of signs and Laguerre's extensions, (Math. Gazette 90, no. 518 (2006), 223-234).
An online copy can be found here.

A: This kind of equation does not show solutions which could be expressed analytically and only numerical methods, such as Newton, should be used.
But before, you could notice that this function varies extremely fast and, in opinion, it should be better to rewrite the equation as $$\log(2011^x+2012^x+2013^x)=\log(2014^x)$$ If there were only one term, it will be the intersection of two straight lines.
Starting with a "reasonable" guess $x_0$, it will be updated according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, let us write $$f(x)=\log(2011^x+2012^x+2013^x)-\log(2014^x)$$  As said, the problem is to start with a reasonable value. So, let us try with $3 \times 2012^x-2014^x=0$ which gives $x_0=\frac{\log (3)}{\log \left(\frac{1007}{1006}\right)} \simeq 1105.75$ and now apply Newton.
The successive iterates are then : $1225.43$, $1226.69$  which is the solution for six significant figures.
Using this transform makes the convergence very fast. Let us try with $x_0=0$; the iterates of Newton method are :$1105.66$, $1225.43$, $1226.69$. 
As you can see, all the trick is the switch to logarithms.
A: Starting equation is equivalent to this one:
$$
\left(\frac{2011}{2014}\right)^x +
\left(\frac{2012}{2014}\right)^x +
\left(\frac{2013}{2014}\right)^x =1.
$$
Denote $a=\frac{2011}{2014}, b=\frac{2012}{2014}, c=\frac{2013}{2014}$.
Function $f(x)=a^x+b^x+c^x-1$ is continuous and decreasing (since $\ln a$, $\ln b$,$\ln c$ are negative):
$$f'(x) =  a^x\ln a + b^x \ln b+c^x\ln c<0.$$
$f(0)=2$, $f(+\infty)=-1$, so there is only $1$ solution.
