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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CAT 2020 Quant questions with Solutions