Question

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

Answer 1

$$\begin{gathered} \because 2^{x+y}=2^{x-y}=\sqrt{8} \\ \therefore x+y=x-y \\ \Rightarrow y=0 \end{gathered}$$
Hence, the correct answer is option (c).