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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 ...

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  • $\begingroup$ There is no closed form for the solution of $64 n \log_2 n = 8 n^2$, but one can write the solution using the special Lambert W function, or solve it numerically, e.g, using Newton's Method. One can see by hand that the solution is greater than $2^5$ but less than $2^6$. $\endgroup$ – Travis Willse Feb 22 '16 at 20:17
  • $\begingroup$ @Travis and I am really trying hard to learn math , which I was never good at it . can you provide a more detailed answer? $\endgroup$ – shayan Feb 22 '16 at 20:19
  • $\begingroup$ This equation has two solutions, namely $n = 1.0999970302376...$ and $n=43.55926043688...$, but as Travis said they cannot be expressed by means of the most common elementary functions. In this case one can employ one of the several numerical methods for finding an approximate solution of a generic equation. $\endgroup$ – Giovanni Resta Feb 22 '16 at 20:27
  • $\begingroup$ @shayan In short there's no way to do this by hand, and for most purposes one has to content themselves with numerical approximations. $\endgroup$ – Travis Willse Feb 22 '16 at 21:04
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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.

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