If $2^{x+y}=2^{x-y}=\sqrt{8}$ then the value of $\mathrm{y}$ is
(a) $\frac{1}{2}$
(b) $\frac{3}{2}$
(c) 0
(d) none of these

Answer: (e) 0

Solution:
$\begin{array}{l}
\because 2^{x+y}=2^{x-y}=\sqrt{8} \\
\therefore \mathrm{x}+\mathrm{y}=\mathrm{x}-\mathrm{y} \\
\Rightarrow \mathrm{y}=0
\end{array}$