Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

Believe it or not, this isn't homework. It's been many years since grade school, and I'm trying to brush up on these things. But my intuition isn't helping me here.

share|improve this question
1  
$n\log n = C$ is equivalent to $n^n = e^C$ which has no closed form solution for $n$. There is a solution for $n$ in terms for Lambert W function. – user2468 Aug 30 '12 at 2:26

3 Answers

up vote 1 down vote accepted

If what you want is to solve for $n$, there is no simple way. The solution has $n=e^{W(C)} = \frac C{W(C)}$ where $W()$ is the Lambert W function.

share|improve this answer
up vote 1 down vote

I will take the base of the logarithm to be $e$. (If you are in a different base; just replace the $e$.)

$$n\log n = C$$

$$\log n = \frac{C}{n}$$

$$e^{\frac{C}{n}} = n$$

$$(e^C)^{\frac{1}{n}} = n$$

$$e^C = n^n$$

share|improve this answer
If you want to do that, would it not be simpler to just raise $e$ to both side of the original equation, obtaining $e^{n\log n} = e^C$, then $\big(e^{\log n}\big)^n = e^C$, then $n^n = e^C$? – MJD Aug 30 '12 at 2:22
@MJD Easier is subjective! – Peter Tamaroff Aug 30 '12 at 2:24
@MJD I see. You use $e^{\log n} = n$, and I used $\log x = y$ means $e^y = x$. – William Aug 30 '12 at 2:25
I would have characterized the difference this way: You divided through by $n$ at the beginning, and then multiplied by $n$ again in the last step. I was puzzled that you included these two steps, which cancel out and seemingly accomplish nothing. – MJD Aug 30 '12 at 13:42
up vote 1 down vote

Since $n \ln n = \ln n^n$, just raise $e$ to each side and you get $n^n = e^C$.

If you want to solve for $n$, I don't know of any method that will work (usually in Computer Science we use approximations or we solve it numerically).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.