How to solve $\ln x^n=0$? Consider the equation $\ln x^n=0$ , where $n$ is any positive integer.
I use two method to solve it, but it gives different answer.
Method 1:
$$\ln x^n=0$$
$$x^n=1$$
By De Moivre's Theorem, we get
$$x=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}$$ 
where $ k= 0, 1 ,2 ,3 ,... , n-1$
So there are total $n$ roots in this equation.
Method 2:
Using $ \ln a^b = b \ln a$,
$$\ln x^n=0$$
$$n\ln x=0$$
$$\ln x=0$$
$$x=1$$
Which is only one root.
Which method is correct and what goes wrong?
 A: This is because you are not specifying where you want to solve your equation.
Since $\ln$ is usually defined on $\mathbb {R}_+^*$, you can not accept solutions outside of that set, in your case: $\mathbb C$.
But if you give yourself a definition of the logarithm on $\mathbb C$, then you can also find your complex solutions.

To answer a comment:
If you take this as a definition for a complex logarithm:
$$\log(z)=\ln(z)+i\arg(z),$$
then you get
$$\begin{align*} \log(z^n)=0&\iff \ln\vert z^n\vert+i\arg(z^n)=0 \\ &\iff n\ln\vert z\vert+ni\arg(z)=0\\ &\iff \begin{cases} \vert z\vert =1 \\ n\arg(z)=0 \pmod {2\pi} \end{cases} \\ &\iff  \begin{cases} \vert z\vert =1 \\ \arg(z)=0 \pmod {\frac{2\pi}n}, \end{cases}\end{align*}$$
which gives you the solution you want.
A: One should be sure what definition one is using for the symbol "$\ln$". 


*

*In real analysis, the single-valued real logarithm $\ln(x)$ is only defined for $x>0$; 

*In complex analysis, $\ln(x)$ is usually reserved for positive real number $x$ while the multi-valued complex logarithm is denoted as $\log(z)$ for non-zero complex number $z$. The relation between "$\ln$" and "$\log$" is then given by the definition
$$
\log(z)=\ln(|z|)+i\arg (z),\quad z\in{\bf C}\backslash\{0\}\tag{*}
$$
where
$ \mathrm{arg}(z) := \{ \theta \in {\bf R}: \cos \theta + i \sin \theta = \frac{z}{|z|} \}
$
denotes all the possible arguments of ${z}$ in polar form.

*Of course you could see that some people insists on using $\ln(z)$ for complex number $z$ as well, which would cause unnecessary confusion regarding what the value of $\ln(z)$ should be when $z$ is a positive real number. 


Now, let's stick to the definition and also the symbols in (*), and see what should be the solution to $\log (z^n)=0$. 
By definition, $$\ln(|z^n|)+i\arg(z^n)=0$$
which implies that $|z^n|=1$ and $\arg(z^n)=0$. It means that $z^n=1$ and now you can use the De Moivre's formula. This is essentially what you did in Method 1. 
The problem in Method 2 is that the identity $\ln a^b=b\ln a$ $(a,b>0)$ is only true for the real logarithm, which would a false assumption if you are in the complex world. For instance, it is an instructive exercise to check by definition of the complex logarithm that
$$
\log (-1)^2\neq 2\log(-1). 
$$
A: While the problems with the complex logarithm have already been talked about in the other answers, you also made the mistake of just using the rule $\ln(a^b)=b\cdot \ln(a)$ without thinking about when this is actually true.
When dealing with complex numbers, this rule doesn't hold anymore. The same goes for other rules such as $\log(z_1\cdot z_2)\stackrel{?}{=}\log(z_1)\cdot\log(z_2)$ where $\log$ is the principal branch of the complex logarithm and $z_1,z_2\in\mathbb C$; in general this equality won't hold (see also answer by Jack). I won't talk about complex solutions anymore and instead focus on the equation assuming we are working only with real numbers.
If you are now dealing with the real valued logarithm $\ln$, even then you have to think before just applying any rules. In
$$\ln(x^n)=0$$
you have two variables that can have an impact on the equation. 
The first thing one should remember, is where $\ln(x^n)$ is defined. This gives us $x^n>0$. For now let us just assume that we have $x^n>0$ so that we can try to solve the equation. By applying the rule we get
$$\ln(x^n)=0 \Leftrightarrow n\cdot\ln(x)=0.$$
Is this still well-defined? We assumed that $x^n>0$, but that doesn't necessarily imply $x>0$; if $n$ is even we can have $x<0$ which will still result in $x^n$ being positive. So we now have to distinguish between two cases: "$n$ uneven" and "$n$ even".
If $n$ is uneven, $x^n>0$ implies that $x>0$, therefor both $\ln(x^n)$ and $\ln(x)$ are well-defined and it indeed holds that
$$\ln(x^n)=0 \Leftrightarrow n\cdot \ln(x)=0 \Leftrightarrow x=1.$$
(Again, not talking about complex solutions here)
If $n$ is even, we can't just apply the rule as $x^n>0$ does not imply $x>0$ and $\ln(x)$ with $x<0$ is not well-defined. What we can do instead is using the absolute value $x^n=|x|^n$ which gives us:
$$\ln(x^n)=0 \Leftrightarrow \ln(|x|^n)=0 \Leftrightarrow n\cdot \ln(|x|)=0 \Leftrightarrow |x|=1 \Leftrightarrow x=\pm 1.$$
The main problem with your second solution lies in "just applying some rule". For every rule there are requirements that have to be met; if you don't check for these requirements, mistakes like this can happen very easily.
