CAT 2025 Slot 3 QA Q4 Logarithms & Exponents

Question 4 of the Quantitative Ability (QA) section, CAT 2025 Slot 3, with CAT's official answer and a worked solution.

If \( \left( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } \right) = 25 \) and \( x > 0 \), then the value of \( \left( x ^ { 7 } + \frac { 1 } { x ^ { 7 } } \right) \) is

A
\( 44859 \sqrt { 3 } \)
B
\( 44856 \sqrt { 3 } \)
C
\( 44853 \sqrt { 3 } \)
D
\( 44850 \sqrt { 3 } \)

Answer: (C) 44853 √( 3 ) Logarithms & Exponents MCQ

Solution

\((x + \frac{1}{x})^2 = 25 + 2 = 27\), so \(t = x + \frac{1}{x} = 3\sqrt{3}\) (as \(x > 0\)).
\(x^3 + \frac{1}{x^3} = t^3 - 3t = 81\sqrt{3} - 9\sqrt{3} = 72\sqrt{3}\).
\(x^4 + \frac{1}{x^4} = 25^2 - 2 = 623\).
\(x^7 + \frac{1}{x^7} = (x^3 + \frac{1}{x^3})(x^4 + \frac{1}{x^4}) - (x + \frac{1}{x}) = 72 \times 623\sqrt{3} - 3\sqrt{3} = 44853\sqrt{3}\).

Shortcut. Product of the 3rd and 4th power sums minus the 1st power sum gives the 7th.