7.140. $\left\{\begin{array}{l}4^{x+y}=2^{y-x} \\ 4^{\log _{\sqrt{2}} x}=y^{4}-5\end{array}\right.$
## Solution.

Domain of definition: $x>0$. From the condition

$$
\left\{\begin{array} { l } 
{ 2 ^ { 2 x + 2 y } = 2 ^ { y - x } , } \\
{ 2 ^ { \log _ { 2 } x ^ { 4 } } = y ^ { 4 } - 5 }
\end{array} \Rightarrow \left\{\begin{array}{l}
2 x+2 y=y-x, \\
x^{4}=y^{4}-5
\end{array} \Rightarrow y=-3 x\right.\right.
$$

From the second equation $x^{4}=(-3 x)^{4}-5, x^{4}=\frac{1}{16}$, from which, considering the domain of definition, we get $x=\frac{1}{2}, y=-\frac{3}{2}$.

Answer: $\left(\frac{1}{2} ;-\frac{3}{2}\right)$