CAT 2024 Slot 1 QA Question & Solution
Question
The sum of all real values of k for which $\left(\cfrac{1}{8}\right)^{k}\times \left(\cfrac{1}{32768}\right)^{\cfrac{1}{3}}=\cfrac{1}{8}\times \left(\cfrac{1}{32768}\right)^{\cfrac{1}{k}}$, is
Options
Solution
To solve this question, we need to immediately recognise the fact that, $32768=8^5$
Substituting this in the above given equation,
$\left(\dfrac{1}{8}\right)^k\times\ \left(\dfrac{1}{8}\right)^{5\times\ \dfrac{1}{3}}=\left(\dfrac{1}{8}\right)\times\ \left(\dfrac{1}{8}\right)^{5\times\ \dfrac{1}{k}}$
Since the bases are equal, we can equate the powers on either side of the equation,
$k+\frac{5}{3}=1+\frac{5}{k}$
$\frac{\left(3k+5\right)}{3}=\frac{\left(k+5\right)}{k}$
$3k^2+5k=3k+15$
$3k^2+2k-15=0$
Here in the given quadratic equation, the Discriminant is greater than 0, $2^2-\left(4\right)\left(3\right)\left(-15\right)>0$
That means both the roots are real, hence we can simply take the sum of the roots of the quadratic equation in k,
Which in a standard quadratic equation of the form $ax^2+bx+c$ is $-\frac{b}{a}$
Here, the sum of the real values of k is $-\frac{2}{3}$
