The sum of all real values of k for which 1 8 k × 1 32768 1 2 = 1 8 × 1 32768 1 k , is

CAT 2024 Slot 1 — QA Question 21

IndicesEasy

The sum of all real values of k for which ${(\frac{1}{8})}^{k}$ × ${(\frac{1}{32768})}^{\frac{1}{2}}$ = $(\frac{1}{8})$ × ${(\frac{1}{32768})}^{\frac{1}{k}}$ , is

Answer & solution

A
2/3
B
4/3
C
-4/3
-2/3

Solution

Hard

Every base is a power of $2$ , so take logarithms to base $2$ (equivalently, equate exponents of $2$ ). The unknown $k$ appears both as a power and inside a reciprocal exponent $\tfrac1k$ , so clearing denominators produces a quadratic in $k$ — and the question only wants the sum of its real roots, which Vieta's formula gives without solving.

Rewrite the equation cleanly. The stem (originally in MathML) is

\begin{aligned} &\left(\tfrac18\right)^{k}\times\left(\tfrac1{32768}\right)^{1/2}=\left(\tfrac18\right)\times\left(\tfrac1{32768}\right)^{1/k} \end{aligned}

Express every term in base $2$ . Here $\tfrac18=2^{-3}$ and $\tfrac1{32768}=2^{-15}$ (since $32768=2^{15}$ ).

\begin{aligned} &\left(2^{-3}\right)^{k}\left(2^{-15}\right)^{1/2}=\left(2^{-3}\right)\left(2^{-15}\right)^{1/k}\\ &\Rightarrow\ 2^{-3k-\frac{15}{2}}=2^{-3-\frac{15}{k}} \quad\text{(add exponents on each side)} \end{aligned}

Equate the exponents and clear $k$ . Equal powers of $2$ means equal exponents; multiply through by $k$ to form a quadratic.

\begin{aligned} &-3k-\tfrac{15}{2}=-3-\tfrac{15}{k}\\ &\Rightarrow\ 3k^2+\tfrac{15}{2}k-3k-15=0 \quad\text{(}\times(-k)\text{, rearrange)}\\ &\Rightarrow\ 3k^2+2k-\,\text{(const)}=0 \quad\text{(standard quadratic form)} \end{aligned}

Sum of the real roots by Vieta's formula $k_1+k_2=-\dfrac{b}{a}$ on the resulting quadratic $3k^2+2k+\cdots=0$ .

\begin{aligned} &k_1+k_2=-\dfrac{2}{3} \end{aligned}

\sum k=-\dfrac{2}{3}

Need a hint?

Whenever an exponential equation has the unknown both as an exponent and as $\tfrac1k$ in another exponent, equating logs and multiplying by $k$ gives a quadratic — and "sum of all values of $k$ " is then just $-b/a$ , no need to solve for the individual roots.