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

Question

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 05 '16 at 18:07 — Daniel Fischer, Sep 05 '16 at 18:07
Thank you, this helped me solve the problem. – Clark Bell Sep 05 '16 at 18:22 — Clark Bell, Sep 05 '16 at 18:22

score 0 · Accepted Answer · answered Sep 05 '16 at 18:27

0

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

Can you take it from here?

answered Sep 05 '16 at 18:27

Nathaniel B

1,096

Yes. The answer to the limit above is 15 correct? This is essentially what I did to solve the problem. – Clark Bell Sep 05 '16 at 18:31
Yes, it is. (Filler so I can post this comment) – Nathaniel B Sep 06 '16 at 03:04

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

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