$\begin{array}{l}
{\left( {\dfrac{3}{5}} \right)^x} + \dfrac{{98}}{{125}} = 1\\
{\left( {\dfrac{3}{5}} \right)^x} = 1 - \dfrac{{98}}{{125}}\\
{\left( {\dfrac{3}{5}} \right)^x} = \dfrac{{27}}{{125}}\\
{\left( {\dfrac{3}{5}} \right)^x} = {\left( {\dfrac{3}{5}} \right)^3}\\
\Rightarrow x = 3
\end{array}$