Question 3.

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 $-\frac{2}{3}$