# Exercise 3.2.8

Show that $k\ln{k} = \Theta(n)$ implies $k = \Theta(n/\ln{n})$.

From the symmetry of $\Theta$:

$$k\ln{k} = \Theta(n) \Rightarrow n = \Theta(k\ln{k})$$

Lets find $\ln{n}$:

$$\ln{n} = \Theta(\ln(k\ln{k})) = \Theta(\ln{k} + \ln\ln{k}) = \Theta(\ln{k})$$

Let's divide the two:

$$\frac{n}{\ln{n}} = \frac{\Theta(k\ln{k})}{\Theta(\ln{k})} = \Theta{\frac{k\ln{k}}{\ln{k}}} = \Theta(k)$$