CAT 2024 Slot 1 · Quant · Question 3

Both bases are powers of 2:

\frac{1}{8}={2}^{-3} and \frac{1}{32768}={2}^{-15}, since {2}^{15}=32768.

The left side becomes {2}^{-3k}\times{2}^{-5}={2}^{-3k-5}.

The right side becomes {2}^{-3}\times{2}^{-\frac{15}{k}}={2}^{-3-\frac{15}{k}}.

Equal powers of the same base force equal exponents: -3k-5=-3-\frac{15}{k}.

Multiplying through by k, -3{k}^{2}-5k=-3k-15.

Rearranging, 3{k}^{2}+2k-15=0.

Its discriminant is 4+180=184>0, so there are two real roots, and the product of roots is -5\ne 0, so neither is the forbidden value k=0.

By Vieta the sum of roots is -\frac{2}{3}.

Shortcut. The question asks for the sum of the roots, not the roots. Once the quadratic appears, Vieta gives -\frac{b}{a} directly — the irrational roots never need to be computed.