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
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 $$ 0 = 1 + \left( {\frac{3} {4}} \right)^{\,x_{\,0} } \left( {1 - \frac{{\delta x}} {4}} \right) + \left( {\frac{5} {4}} \right)^{\,x_{\,0} } \left( {1 + \frac{{\delta x}} {4}} \right) - \left( {\frac{3} {2}} \right)^{\,x_{\,0} } \left( {1 + \frac{{\delta x}} {2}} \right) $$