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.

  • 1
    $\begingroup$ Yes, you are right. In the first interpretation, the limit diverges. And in the second, $n=79$ is correct. $\endgroup$ – Mark Viola Dec 18 '16 at 16:29
  • $\begingroup$ What exactly is the standard for diverging the ln and sigma? $\endgroup$ – juyoung518 Dec 18 '16 at 16:34
  • 2
    $\begingroup$ Moral lesson (for the problem poser and not the OP): when in doubt, parenthesize. $\endgroup$ – J. M. is a poor mathematician Dec 18 '16 at 16:35
  • $\begingroup$ Thanks everyone. I know that in cases like this, the ln and sigma should be considered apart. But are there any specific standards or rules regarding these kinds of problem? $\endgroup$ – juyoung518 Dec 18 '16 at 16:39

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