why $-\log(x) $ is a valuation like $v_p(x)$?

Question

Why $\log$ is used in the definition of valuation?

I can not understand how $\log(x)$ gives rational value always.

We know that p-adic valuation of $x \in \Bbb Q$ is given by $$|x|_p=p^{-v_p(x)}.$$ We also define valuation to be $-\log(x)$.

Also we know that $v_p(x) \in \Bbb Z \subset \mathbb{Q}$.

My question-

Why $-\log(x) $ is a valuation i.e., why $|x|_p=p^{-\log(x)}$ ?

Answer 1

Let $x\in \mathbb{Q}$ be a rational number. Then by fundamental theorem of arithmetic you can uniquely factorize

$$|x| = \prod_{p\in \mathbb{P}}p^{\nu_p(x)}$$

where $\nu_p(x)\in \mathbb{Z}$. You can define $p$-adic absolute value by $|x|_p = p^{-\nu_p(x)}$. The idea is that sequence $\{x_n\}_{n\in \mathbb{N}}$ of rational numbers is convergent to $0$ if and only if "$x_n$ are more divisible by $p$ as $n$ tends to $+\infty$".

You can also define equivalent absolute value by $|x|'_p=\beta^{-\nu_p(x)}$ for $\beta>1$, but it defines the same topology on $\mathbb{Q}$ (which is precisely why we call it equivalent).

I don't know any definition of $p$-adic absolute value that involves the usual logarithm of some base.

Edit after OP's clarification.

The OP's question is an issue with interpreting problem 28 on page 26 of F. Gouvea's book "p-adic numbers, An Introduction".

Fix a field $K$. The point of the exercise is to show that the map

$$\big\{\mbox{non-archimedean absolute values on }K\big\}\in |-|\mapsto -\log(|x|)\in \big\{\mbox{ discrete valuations on }K\big\}$$

is a bijective correspondence, which is a correct mathematical result.

Note that you missed the fact that logarithm is applied to absolute value in the exercise.

Let me illustrate this by using $p$-adic absolute value.

$$|x|_p = p^{-\nu_p(x)}$$

where $\nu_p$ is $p$-adic valuation (defined in the first part of my answer). Exercise told us to apply the logarithm to an absolute value $|x|_p$ and multiply the result by $(-1)$. Hence we obtain

$$-\log(|x|_p) = -\left(-\nu_p(x)\right)\cdot \log(p) = \log(p)\cdot \nu_p(x)$$

so in this way we recover $p$-adic valuation (up to positive multiplicative constant, which yields equivalent valuation).

On the other hand in the formulation of your question you applied logarithm to $x\in \mathbb{Q}$ and expected

$|x|_p = p^{-\log(x)}$

to be a $p$-adic absolute value, which under any reasonable circumstances does not take place and is not defined for $x<0$.