# Prove that $15n \log(5n) = \Theta(n \log(n))$

I must determine the positive constants $c_1$, $c_2$, and $n_0$ such that

$$c_1 n \log(n) \leq 15n \log(5n) \leq c_2 n \log(n)$$

To simplify I attempted to divide by

$$n \log(n))$$

however this yielded a messy result involving changing the base of the logarithm.

• Recall that $\log (5n) = \log n + \log 5$. – Daniel Fischer Sep 5 '16 at 18:07
• Thank you, this helped me solve the problem. – Clark Bell Sep 5 '16 at 18:22

\begin{align}\lim_{x\to\infty}\frac{15x\log(5x)}{x\log x} &=\lim_{x\to\infty}\frac{15x(\log x+\log 5)}{x\log x}\\ &=\lim_{x\to\infty}\frac{15x\log x+15x\log 5}{x\log x}\\ &=\lim_{x\to\infty}\frac{15x\log x}{x\log x}+\frac{(15\log 5)x}{x\log x} \end{align}