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If $x=\left ( 4096 \right )^{7+4\sqrt{3}}$, then which of the following equals $64$?

  1. $\frac{x^{7}}{x^{2\sqrt{3}}}$
  2. $\frac{x^{7}}{x^{4\sqrt{3}}}$
  3. $\frac{x\frac{7}{2}}{x\frac{4}{\sqrt{3}}}$
  4. $\frac{x\frac{7}{2}}{x^{2\sqrt{3}}}$
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Given that, $ x = (4096)^{7+4 \sqrt{3}} $

$ \Rightarrow x^{\frac{1}{7+4 \sqrt{3}}} = 4096 $

Do the rationalization, we get

$ \Rightarrow x^{ \left( \frac{1}{7+4 \sqrt{3}} \times \frac{7-4\sqrt{3}}{7-4\sqrt{3}} \right)} = 4096 $

$ \Rightarrow x^{ \left( \frac{7-4\sqrt{3}}{49-48} \right)} = 2^{12} \quad [ \because a^{2} – b^{2} = (a+b)(a-b)] $

$ \Rightarrow x^{7-4 \sqrt{3}} = (2^{6})^{2} $

$ \Rightarrow x^{7-4 \sqrt{3}} = (64)^{2} $

$ \Rightarrow x^{\frac{1}{2} (7-4 \sqrt{3})} = 64 $

$ \Rightarrow 64 = x^{\left( \frac{7-4 \sqrt{3}}{2}\right)} $

$ \Rightarrow 64 = x^{ \left(\frac{7}{2} - 2 \sqrt{3} \right)} $

$ \Rightarrow 64 = x^{\frac{7}{2}} \cdot x^{-2 \sqrt{3}} $

$ \Rightarrow \boxed{64 = \frac{x^{\frac{7}{2}}}{x^{2 \sqrt{3}}}} \quad \left[ \because x^{-a} = \dfrac{1}{x^{a}}\right] $

Correct Answer$: \text{D}$
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