[tex]log_{x} x^{y}=log_{x} y^{x}[/tex](x>0 from the second equation) [tex]\Leftrightarrow y=x.log_{x}y[/tex] (1) Similarly from the second equation [tex]log_{x}x^{3}=log_{x}y^{2} \Leftrightarrow log_{x}y=\frac{3}{2}[/tex] (2). From (1) and (2) [tex]y=\frac{3x}{ 2}[/tex]. Then again from the second equation of the system we have [tex]x^{3}=\frac{9}{4 }x^{2}[/tex] and it is easy then...