CAT 2020

Slot 1

QA

Question & Solution

Step 2: Rewrite each option using $x = 2^A$

Option 1: $$ \frac{x^{7}}{x^{2\sqrt{3}}} = x^{7 - 2\sqrt{3}} = 2^{A(7 - 2\sqrt{3})} $$

Option 2: $$ \frac{x^{7/2}}{x^{4/\sqrt{3}}} = x^{\frac{7}{2} - \frac{4}{\sqrt{3}}} = 2^{A\left(\frac{7}{2} - \frac{4}{\sqrt{3}}\right)} $$

Option 3: $$ \frac{x^{7/2}}{x^{2\sqrt{3}}} = x^{\frac{7}{2} - 2\sqrt{3}} = 2^{A\left(\frac{7}{2} - 2\sqrt{3}\right)} $$

Option 4: $$ \frac{x^{7}}{x^{4\sqrt{3}}} = x^{7 - 4\sqrt{3}} = 2^{A(7 - 4\sqrt{3})} $$

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Step 3: We want the expression to equal 64

Since: $$ 64 = 2^{6} $$

We need: $$ A \cdot (\text{exponent difference}) = 6 $$

Recall: $$ A = 12(7 + 4\sqrt{3}) $$

So we want: $$ 12(7 + 4\sqrt{3}) \cdot k = 6 $$

Thus: $$ k = \frac{6}{12(7 + 4\sqrt{3})} = \frac{1}{2(7 + 4\sqrt{3})} $$

We check which option has: $$ k = \frac{7}{2} - 2\sqrt{3},\quad 7 - 2\sqrt{3},\quad \frac{7}{2} - \frac{4}{\sqrt{3}},\quad 7 - 4\sqrt{3} $$

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Step 4: Test Option 3 (most likely candidate)

Option 3 exponent: $$ k = \frac{7}{2} - 2\sqrt{3} $$

Multiply with $A$: $$ 12(7 + 4\sqrt{3})\left(\frac{7}{2} - 2\sqrt{3}\right) $$

Factor out 6: $$ = 6(7 + 4\sqrt{3})(7 - 4\sqrt{3}) $$

Since: $$ (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 49 - 48 = 1 $$

Thus: $$ A \cdot k = 6(1) = 6 $$

Therefore: $$ 2^{A k} = 2^{6} = 64 $$

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Final Answer: Option C equals $\boxed{64}$.