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

The correct option is (c): $0$

Given: $2^{x+y}=2^{x-y}=\sqrt{8}$

$\because \sqrt{8}=\sqrt{2\times 2\times 2}=\sqrt{2}\times \sqrt{2}\times \sqrt{2}=(\sqrt{2}{)}^3=2^{\frac{1}{2}\times 3}=2^{\frac{3}{2}}$

$2^{x+y}=2^{x-y}=2^{\frac{3}{2}}$

So, $2^{x+y}=2^{\frac{3}{2}}$

$\Rightarrow x+y=\frac{3}{2}$

$\Rightarrow 2x+2y=3\dots \dots \left(i\right)$

And, $2^{x-y}=2^{\frac{3}{2}}$

$\Rightarrow x-y=\frac{3}{2}$

$\Rightarrow 2x-2y=3\dots \dots \left(i\right)$

On adding eq.(i) and (ii), we get

$4x=6$

$\Rightarrow x=\frac{6}{4}=\frac{3}{2}$

Now from eq.(i), we have

$2\times \frac{3}{2}+2y=3$

$\Rightarrow 2y=0$

$\Rightarrow y=0$