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

\binom{x^{\frac{7}{2}}}/{x^{\frac{4}\sqrt{3}}}

\binom{x^{\frac{7}{2}}}/{x^{2 \sqrt{3}}

CAT 2020 Exam Shift 1 Solved Paper

Question : 67 of 76

Marks: +1, -0

x=(4096)^{7+√{3}}

, then which of the following equals 64 ?

Solution:

x=(4096)^{7+4 √{3}}

x^{\;{1}/{7+4 √{3}}}=(4096)

On rationalizing

7+4 √{3}

, we get

\;{1}/{7+4 √{3}}=7-4 √{3}

∴ x^{7-4 √{3}}=(64)^{2}

∴ 64=x^{\;{7-4 √{3}}/{2}}

64=\;{x^{{7}/{2}}}/{x^{2 √{3}}}

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