Does $P(\exp_p(a),\exp_p(b))=0$ imply $P=0$, where $\exp_p(\cdot)$ is $p$-adic exponential?

Question

Classical case:

Let $\{a,b\}$ be linearly independent set over $\mathbb Q$ and $\{e^{at},e^{bt}\}$ be linearly independent set over $\mathbb Q[[t]]$. Suppose $P(x,y)$ is a polynomial over $\mathbb Q$. Then it seems $P(e^{a},e^{b})=0$ implies $P=0$, because if $P(x,y)=Ax^2+Bxy+Cy^2 \in \mathbb Q[x,y]$. Then, \begin{align}& P(e^{a},e^{b})=0 \\ \Rightarrow & Ae^{2a}+Be^{a+b}+Ce^{2b}=0 \\ \Rightarrow &A=B=C=0,~\text{because $a,b$ are linearly independent} \end{align} The same is true for any arbitrary polynomials $P(x,y) \in \mathbb Q[x,y]$, because $P(e^{a},e^{b})$ can be written as linear combinations of exponentials of the form $e^{c}$, where $c=n_1a+n_2b$, $n_1,n_2 \in \mathbb N$. But since $\{a,b\}$ is linearly independent, $P=0$.

Does the same hold for $p$-adic exponential as well ?

$p$-adic case:

Let $\exp_p$ denotes the $p$-adic exponential function satisfying $\exp_p(x+y)=\exp_p(x) \exp_p(y)$ for $|x,y|_p<p^{-\frac{1}{p-1}}$.

Let $\{a,b\}$ be linearly independent set over $\mathbb Q$ and $\{\exp_p(at), \exp_p(bt)\}$ be linearly independent set over $\mathbb Q[[t]]$. Suppose $P(x,y)$ is a polynomial over $\mathbb Q$. Then it seems $P(\exp_p(a), \exp_p(b))=0$ implies $P=0$, because if $P(x,y)=Ax^2+Bxy+Cy^2 \in \mathbb Q[x,y]$. Then, \begin{align}& P(\exp_p(a), \exp_p(b))=0 \\ \Rightarrow & A\exp_p(2a)+B \exp_p(a+b)+C\exp_p(2b)=0 \end{align} I think, as $a,b$ are linearly independent, the $p$-adic exponential $\exp_p(2a),~\exp_p(2b)$ and $\exp_p(2(a+b))$ will not cancel, so we must have $A=B=C=0$. In other word, $P=0$. The same will be true for arbitrary polynomial.

So the answer seems to be positive, if there is no flaw in my argument.

I appreciate your comments.

Answer 1

As written, both the classical and $p$-adic statements are false. Let $a=\log(9)$ and $b=\log(17)$ (classical or $2$-adic log). Then they are linearly independent over $\mathbb{Q} $ since otherwise we would find integers $n, m$ with $9^n=17^m$. But $P(\exp(a), \exp(b))=0$ for the nontrivial polynomial $P(x, y) = 17x-9y$. (Note that the power series $\exp(at), \exp(bt)$ are still linearly independent as asked in the question.)