Finding roots of $3^x+4^x+5^x-6^x=0$. 
How many real roots of $3^x+4^x+5^x-6^x=0$ exist?

Could anyone please tell both the graphical and a non graphical way?
For graphical way
I am not even able to find critical points.
Could just find one root by hit and trial that is $3$.
Answer is given to be one real root.
thanks in advance
 A: As indicated in Joey's comment, the function is monotonically decreasing from $\infty$ to $-1$, and so there is only
one solution, the $x=3$ you already found.
However, if the input data were different it might have been difficult to estimate a solution, around which to graph or start an iteration.
Therefore let me take this as an example to expose a possible approach that
leads to a polynomial approximation that might be useful to find a first "location" of the zero.
$$
\begin{gathered}
  1 = \left( {\frac{{4 - 1}}
{6}} \right)^{\,x}  + \left( {\frac{4}
{6}} \right)^{\,x}  + \left( {\frac{{4 + 1}}
{6}} \right)^{\,x}  \hfill \\
  \left( {1 + \frac{1}
{2}} \right)^{\,x}  = 1 + \left( {1 - \frac{1}
{4}} \right)^{\,x}  + \left( {1 + \frac{1}
{4}} \right)^{\,x}  = 1 + \left( {1 + \frac{1}
{4}} \right)^{\,x} \left( {1 + \left( {\frac{3}
{5}} \right)^{\,x} } \right) \hfill \\
  \sum\limits_{0\, \leqslant \,k} {\left( \begin{gathered}
  x \\ 
  k \\ 
\end{gathered}  \right)\left( {\frac{1}
{2}} \right)^{\,k} }  = 1 + \sum\limits_{0\, \leqslant \,k} {\left( \begin{gathered}
  x \\ 
  k \\ 
\end{gathered}  \right)\left( {1 + \left( { - 1} \right)^{\,k} } \right)\left( {\frac{1}
{4}} \right)^{\,k} }  = 1 + 2\sum\limits_{0\, \leqslant \,k} {\left( \begin{gathered}
  x \\ 
  2k \\ 
\end{gathered}  \right)\left( {\frac{1}
{4}} \right)^{\,2k} }  \hfill \\ 
\end{gathered} 
$$
Finally note that the second expression can be easily rewritten around the 
approximation found ($x_0$)
$$
\left( {\frac{3}
{2}} \right)^{\,x_{\,0} } \left( {1 + \frac{1}
{2}} \right)^{\,x - x_{\,0} }  = 1 + \left( {\frac{3}
{4}} \right)^{\,x_{\,0} } \left( {1 - \frac{1}
{4}} \right)^{\,x - x_{\,0} }  + \left( {\frac{5}
{4}} \right)^{\,x_{\,0} } \left( {1 + \frac{1}
{4}} \right)^{\,x - x_{\,0} } 
$$
and substituting the binomials with their first order development we
obtain a linear recursion in ${\delta x}$, which corresponds to 
the tangent method
$$
0 = 1 + \left( {\frac{3}
{4}} \right)^{\,x_{\,0} } \left( {1 - \frac{{\delta x}}
{4}} \right) + \left( {\frac{5}
{4}} \right)^{\,x_{\,0} } \left( {1 + \frac{{\delta x}}
{4}} \right) - \left( {\frac{3}
{2}} \right)^{\,x_{\,0} } \left( {1 + \frac{{\delta x}}
{2}} \right)
$$
