calculating logarithmic equation: $64n\log_2(n)<8n^2$

Question

I have an equation , which is like this $64n\log n < 8n^2$ . (the base of logarithm is 2) I know how to solve the logarithmic equations . I am a programmer , so I wrote a simple program and answer is 44 . my tries , all end up here : $8\log_2(n) < n \Longrightarrow n^8 < 2^n \Longrightarrow n^8 - 2^n < 0$ , and here is where I can't solve the problem ! I think I am too close , but does not have any clue how to reach 44 ...

Answer 1

I'm sorry to disappoint you, but one cannot solve an inequality with complex numbers, as my solution will most likely be:

$$64n\log_2(n)=8n^2$$

$$8\frac{\ln(n)}{\ln(2)}=n$$

$$\ln(n)=\frac{n\ln(2)}{8}$$

$$n=e^{\frac{n\ln(2)}{8}}$$

$$ne^{-\frac{n\ln(2)}{8}}=1$$

Multiply both sides by $-\frac{\ln(2)}{8}$:

$$-\frac{n\ln(2)}{8}e^{-\frac{n\ln(2)}{8}}=-\frac{\ln(2)}{8}$$

$$-\frac{n\ln(2)}{8}=W_k\left(-\frac{\ln(2)}{8}\right)$$

$$n=\frac{W_k\left(-\frac{\ln(2)}{8}\right)}{-\frac{\ln(2)}{8}}=e^{-W_k\left(-\frac{\ln(2)}{8}\right)}$$

If we have $k=0$, the primary branch, we may get a real solution. Other branches will most likely be complex.