[tex]0.2=\frac{2}{10}=\frac{1}{5}=5^{-1}\ \ \ |used\ a^{-1}=\frac{1}{a}\ for\ a\neq0\\\\25=5^2\\\\therefore\\\\0.2^x=25\to(5^{-1})^x=5^2\ \ \ |use\ (a^n)^m=a^{n\cdot m}\\\\5^{-1(x)}=5^2\\\\5^{-x}=5^2\iff-x=2\to\boxed{x=-2}\\\\check:\\\\0.2^x=0.2^{-2}=\left(\frac{1}{5}\right)^{-2}=5^2=25\ O.K.\ :)[/tex]
