CAT 2020 Slot 1 Question Paper Question 64

Question 64

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

Solution

$$x=2^{12\left(7+4\sqrt{\ 3}\right)}$$.

$$x^{\frac{7}{2}}=2^{42\left(7+4\sqrt{\ 3}\right)}$$

$$x^{2\sqrt{\ 3}}=2^{24\sqrt{\ 3}\left(7+4\sqrt{\ 3}\right)}$$

$$\frac{x^{\frac{7}{2}}}{x^{2{\sqrt{3}}}}$$ = $$2^{\left(7+4\sqrt{\ 3}\right)\left(42-24\sqrt{\ 3}\right)}=2^{\left(7+4\sqrt{\ 3}\right)\left(7-4\sqrt{\ 3}\right)6}$$ =$$2^6$$.

Hence C is correct answer.


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