How to calculate $\log_p x$ in p-adic analysis? I'm studying p-adic analysis and recently I've learned about the p-adic logarithm function but I can't understand very well how the process of calculating the value should be done.
As an exercise I'm trying to find the 7-adic expansion of $\log_7 42$.
From the definition I know that in order to calculate $\log_p x$ for an arbitrary $x\in\Omega$, I have to write $x=p^r\omega(x_1)\langle x_1\rangle$ where $r=ord(x)=\frac{a}{b}$ and $p^r$ is formally a root of $x^b-p^a=0$, $\omega(x_1)$ is a $(p^f-1)$-th root of 1, and $\langle x_1\rangle$ is in the disc around 1 with radius 1, that is $|\langle x_1\rangle-1|_p<1$. Then 
$$\log_p x=\sum\frac{(-1)^{n+1}(\langle x_1\rangle-1)^n}{n}.$$
Now, I have that $42=7(6)$ where $7$ is a root of $x-7=0$, the problem I facing now is that I don know how to calculate the adequate root of 1. I mean, which root to choose? Can I choose an arbitrary $f$? Like for example $f=1$ which gives me that $(-1)^{7^1-1}=1$ that is $-1$ is a 6-th root of 1, and after plugging this into the equation I get $42=7(-1)(-6)$ and $\langle x_1\rangle=-6$, hence  
$$\log_p 42=\sum\frac{(-1)^{n+1}(-6-1)^n}{n}=-\sum\frac{7^n}{n}$$
I'm not sure that my answer is correct but I've checked the solution on my book and this gives the correct answer at least for the first 4 digits of the 7-adic expansion (the solution asks precisely for the first 4 digits).
I would like to know if the process is correct for the particular problem, but also what should be the approach for the general case. 
I have a suspicion that my solution should be correct, because $f\cdot e=n$ where n is the degree of the extension of $\mathbb{Q}_p$ which contains $x$ and $e$ is the index of ramification. In my case (I'm not sure) I have $e=1$ and $n=1$ since $42\in\mathbb{Q}_p$, hence $f=1$ which is preceisely what I've done.
It will be great to have a practical clarification of what $e$ and $f$ could  mean for a specific problem like this (more than just the index of ramification and the residue field degree).
 A: The root of unity that you choose is the unique one that’s congruent to your unit modulo the maximal ideal. In your case of $42$, your unit is $6$, and the unique root of unity that’s congruent to it is $-1$. So your calculation is correct. It would have been much more fun if you had chosen $35=7\cdot5$; then  you would have needed to find the sixth root of unity that’s congruent to $5$. You may do this computationally without Hensel just by applying $z\mapsto z^7$ repeatedly, starting with $z_0=5$. You get one more $7$-adic place of accuracy in each iteration, so it’s not as fast as Newton-Raphson.
In general if you have a finite extension $k\supset\Bbb Q_p$, then since the $\Bbb Q_p$-absolute value $|\star|_p$ extends uniquely to $k$, you form the set of all elements $z\in k$ with $|z|_p\le1$. This is the ring of integers of $k$, call it $\mathfrak o$. It has the unique maximal ideal $\mathfrak m$, consisting of all $z\in k$ with $|z|_p<1$. The field $\kappa=\mathfrak o/\mathfrak m$ is the residue field belonging to $k$, and it’s a finite extension of the corresponding construct for $\Bbb Q_p$, which is $\Bbb Z/p\Bbb Z=\Bbb F_p$. The field degree $[\kappa:\Bbb F_p]$ is your $f$. In particular, if $k=\Bbb Q_p$, $f=1$, as in your case.
The multiplicative subgroup of the reals, $|k^\times|_p$, of all the absolute values of nonzero elements of $k$, contains the corresponding group for $\Bbb Q_p$, which is the set of all powers of $p$, you might write it $p^{\Bbb Z}$. You can show easily enough that the index $(|K^\times|_p:p^{\Bbb Z})$ is finite; this is the ramification index, $e$. Once more, if $k=\Bbb Q_p$, then $e=1$. Always, no matter what, if $[k:\Bbb Q_p]=n$, you have the nice relation $ef=n$.
Examples: Let’s see what happens when you adjoin a cube root of unity to $\Bbb Q_p$ for all possible $p$’s. The $\Bbb Q$-minimal polynomial of a primitive cube root of $1$ is $\Phi_3(X)=X^2+X+1$. This is irreducible over $\Bbb Q_2$, and I hope you know that in characteristic two, the roots of this polynomial generate the field with $4$ elements. So $e=1$, $f=2$ here. For $p=3$, since the roots of $\Phi_3$ are $-\frac12\pm\frac{\sqrt{-3}}2$, you’re adjoining the square root of $-3$, which necessarily has absolute value $1/\sqrt3$, so you have $e=2$, $f=1$. This is the only $p$ for which $e\ne1$. For all other primes $p\ge5$, it’s just a matter of the congruence of $p$ modulo $3$, since if $p\equiv1\pmod3$, then $p-1$, the number of nonzero elements of $\Bbb F_p^\times$, is divisible by $3$, so that there already are cube roots of unity in $\Bbb Q_p$. You can calculate them by the method I mentioned above, but Newton-Raphson works far faster. On the other hand, if $p\equiv-1\pmod3$, then $3$ does not divide the number of nonzero elements of $\Bbb F_p$, and you need to make a quadratic extension to get them. Thus in this case, $f=2$, $e=1$.
