${\log ^x} = {\log ^{\sqrt ۲ }} + \frac{۱}{۲}{\log ^{\frac{۲}{{۲۵}}}}$، مقدار $x$ کدام است؟
${\log ^x} = {\log ^{\sqrt 2 }} + \frac{1}{2}{\log ^{\frac{2}{{25}}}}$$ \Rightarrow {\log ^x} = {\log ^{\sqrt 2 }} + {\log ^{{{\left( {\frac{2}{{25}}} \right)}^{\frac{1}{2}}}}}$${\log ^x} = {\log ^{\sqrt 2 }} + {\log ^{\sqrt {\frac{2}{{25}}} }}$$ \Rightarrow {\log ^x} = {\log ^{\sqrt 2 \times \frac{{\sqrt 2 }}{5}}}$$ \Rightarrow {\log ^x} = {\log ^{\frac{2}{5}}} \Rightarrow x = \frac{2}{5} = 0/4$