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# CAT 2020 Quant Question [Slot 1] with Solution 17

Question

If $x = {(4096)^{7 + 4\sqrt 3 }}$, 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. $\frac{{{x^{\frac{7}{2}}}}}{{{x^{2\sqrt 3 }}}}$
Option: 4
Solution:

$x = {(4096)^{7 + 4\sqrt 3 }}$

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

On rationalizing $7 + 4\sqrt 3$, we get $\frac{1}{{7 + 4\sqrt 3 }} = 7 - 4\sqrt 3$

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

$\therefore 64 = {x^{\frac{{7 - 4\sqrt 3 }}{2}}}$

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

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