The original problem is this: $$\lim_{x\to 0}\frac{1}{2x}\ln \frac{1}{n}\sum_1^n {e^{kx}} = 20$$
Find out the value of $n$, which is a natural number.
But I'm having a bit of trouble solving this problem.
My problem is that I think that this question can be interpreted in two ways. Since there is no brackets limiting the range of the $\ln$, I believe it can be interpreted as both :
1) $$ \lim_{x\to 0}\frac{1}{2x}\ln \left(\frac{1}{n}\right)\sum_1^n {e^{kx}} = 20 $$
2) $$ \lim_{x\to 0}\frac{1}{2x}\ln \left(\frac{1}{n}\sum_1^n {e^{kx}}\right) = 20 $$
If I interpret the problem as 2), the value of $n$ becomes $79$. If I interpret the problem as 1), I can't get the value of $n$.
Am I right in assuming that all two of my interpretations are correct?
What I mean is, just like $\cos x \cos x$ is different from $\cos(x \cos x)$, shouldn't the range of the ln be specified using brackets? Without these brackets, can this equation be interpreted in two ways, just like I did?
And if my first interpretation is correct, could anyone please explain in detail why ln and sigma can be diverged? I certainly learned it at some point at school, but I don't exactly remember the specific details.