Find x of `4^{\frac{x}{y}+\frac{y}{x}}=32`
Find x , if \displaystyle \begin{array}{|l}4^{\frac{x}{y}+\frac{y}{x}}=32\\ \log_3(x-y)+\log_3(x+y)=1\end{array} 4 y x + x y = 32 lo g 3 ( x − y ) + lo g 3 ( x + y ) = 1 Solution: Checking if the system is defined for two variables is a hard task, so we shall find the eventual solutions to the system and check directly if the system is defined for them. We shall only write \displaystyle \begin{array}{|l}x+y>0\\x-y>0\end{array} x + y > 0 x − y > 0 for now. \displaystyle \begin{array}{|l}\frac{x}{y}+\frac{y}{x}=log_432\\log_3(x^2-y^2)=1\end{array} y x + x y = l o g 4 32 l o g 3 ( x 2 − y 2 ) = 1 \displaystyle \begin{array}{|l}\frac{x}{y}+\frac{y}{x}=\frac{1}{2}log_232\\x^2-y^2=3\end{array} y x + x y = 2 1 l o g 2 32 x 2 − y 2 = 3 \displaystyle \begin{array}{|l}\frac{x^2+y^2}{xy}=\frac{5}{2}\\x^2-y^2=3\end{array} x y x 2 + y 2 = 2 5 x 2 − y 2 = 3 \displaystyle \begin{array}{|l}x^2+y^2=\frac{5}{2}xy\\x^2-y^2=3\end{a...