MCQ
If $2^{\text{x+y}}=2^{\text{x}-\text{y}}=\sqrt8$ then the value of $y$ is :
- A$\frac{1}{2}$
- B$\frac{3}{2}$
- ✓$0$
- DNone of these.
✓
Answer
Correct option: C.
$0$
$\Rightarrow2\text{x} +\text{y}=2\text{x}-\text{y}=\sqrt{8}$
$\Rightarrow2\text{x}+\text{y}=\sqrt{8}\ ...(\text{i})$
$\Rightarrow2\text{x}-\text{y}=\sqrt{8}\ ...(\text{ii})$
Adding $(i)$ and $(ii),$ we get
$4\text{x}=2\sqrt8$
$\Rightarrow\text{x}=\frac{\sqrt8}{2}$
Substituting $\text{x}=\frac{\sqrt8}{2}$ in $(i),$
we get $y = 0.$
$\Rightarrow2\text{x}+\text{y}=\sqrt{8}\ ...(\text{i})$
$\Rightarrow2\text{x}-\text{y}=\sqrt{8}\ ...(\text{ii})$
Adding $(i)$ and $(ii),$ we get
$4\text{x}=2\sqrt8$
$\Rightarrow\text{x}=\frac{\sqrt8}{2}$
Substituting $\text{x}=\frac{\sqrt8}{2}$ in $(i),$
we get $y = 0.$
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