Question

In the expression $y^{(\sqrt{2}+\sqrt{6})}=\frac{1}{\sqrt[4]{4{11}^{(\sqrt{2}-\sqrt{6})}}},$ find the value of $y.$

• A. $y={11}^{\frac{2-\sqrt{3}}{4}}$
• B. $y={11}^{\frac{4-\sqrt{3}}{4}}$
• C. $y={11}^{\frac{2+\sqrt{3}}{4}}$

Answer 1

The correct option is A $y={11}^{\frac{2-\sqrt{3}}{4}}$

Let's simplify the given expression.

$y^{(\sqrt{2}+\sqrt{6})}=\frac{1}{\sqrt[4]{4{11}^{(\sqrt{2}-\sqrt{6})}}}$

$\Rightarrow y^{(\sqrt{2}+\sqrt{6})}=\frac{1}{{11}^{\frac{(\sqrt{2}-\sqrt{6})}{4}}}$

We know: $\frac{1}{a^x}=a^{-x}$

$\Rightarrow y^{(\sqrt{2}+\sqrt{6})}={11}^{-\frac{(\sqrt{2}-\sqrt{6})}{4}}$

(taking $(\sqrt{2}+\sqrt{6}{)}^{th}$ root in both side of the equation)

$\Rightarrow y^{\frac{(\sqrt{2}+\sqrt{6})}{(\sqrt{2}+\sqrt{6})}}={11}^{-\frac{(\sqrt{2}-\sqrt{6})}{4(\sqrt{2}+\sqrt{6})}}$

$\Rightarrow y={11}^{-\frac{(\sqrt{2}-\sqrt{6})}{4(\sqrt{2}+\sqrt{6})}}$

Let's rationalize denominator of the exponent part.
$-\frac{(\sqrt{2}-\sqrt{6})}{4(\sqrt{2}+\sqrt{6})}=-\frac{(\sqrt{2}-\sqrt{6})}{4(\sqrt{2}+\sqrt{6})}\times \frac{(\sqrt{2}-\sqrt{6})}{(\sqrt{2}-\sqrt{6})}=-\frac{(\sqrt{2}-\sqrt{6}{)}^2}{4\left(\right(\sqrt{2}{)}^2-\left(\sqrt{6}{)}^2\right)}=-\frac{2+6-2\sqrt{2}\times \sqrt{6}}{4(2-6)}=-\frac{8-4\sqrt{3}}{(-16)}=\frac{8-4\sqrt{3}}{16}=\frac{2-\sqrt{3}}{4}$

Substituting this value in above expression,
$y={11}^{\frac{2-\sqrt{3}}{4}}$