Sum of Real Values of Exponential Equation

CAT 2024 Slot 1 · Quantitative Ability · Medium · Exponents and Logarithms

The sum of all real values of kk for which (18)k×(132768)13=18×(132768)1k\left(\frac{1}{8}\right)^k \times \left(\frac{1}{32768}\right)^{\frac{1}{3}} = \frac{1}{8} \times \left(\frac{1}{32768}\right)^{\frac{1}{k}} is

  1. A.

    \frac{2}{3}

  2. B.

    -\frac{2}{3}

  3. C.

    \frac{4}{3}

  4. D.

    -\frac{4}{3}

Answer

B

Explanation

Express the base terms as powers of 22: 18=2−3\frac{1}{8} = 2^{-3} and 132768=2−15\frac{1}{32768} = 2^{-15}.

Substitute into the equation: (2−3)k×(2−15)1/3=2−3×(2−15)1/k(2^{-3})^k \times (2^{-15})^{1/3} = 2^{-3} \times (2^{-15})^{1/k} 2−3k−5=2−3−15k2^{-3k - 5} = 2^{-3 - \frac{15}{k}}

Equating exponents: −3k−5=−3−15k-3k - 5 = -3 - \frac{15}{k} 3k+2=15k3k + 2 = \frac{15}{k} 3k2+2k−15=03k^2 + 2k - 15 = 0

For a quadratic equation 3k2+2k−15=03k^2 + 2k - 15 = 0, the discriminant D=4−4(3)(−15)=184>0D = 4 - 4(3)(-15) = 184 > 0, so both roots are real. Sum of real roots = −ba=−23-\frac{b}{a} = -\frac{2}{3}.

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