# Linear independence over Q of logarithmic powers of prime numbers

I denote $p_k$ the $k^{th}$ prime number ($p_1=2$, etc...)

Clearly, for any $n\in \mathbb{N}^*$, $(\log p_k)_{1\leq k\leq n}$ is linearly independent over $\mathbb{Q}$.

My question concerns a possible generalization: is it true that for any $q,n\in \mathbb{N}^*$, $((\log{p_k})^q)_{1\leq k\leq n}$ is linearly independent over $\mathbb{Q}$ ?

I am of course interested in weaker statements : $q=2$, "for any $n$ there exists $q\geq 2$ such that..." or "there exists $q\geq 2$ such that for any $n$...".

I am not an expert in the field, sorry if this is actually trivial.

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Schanuel's conjecture (en.wikipedia.org/wiki/Schanuel's_conjecture) implies the stronger statement that the logarithms of the primes are algebraically independent. I think statements like these are generally expected to be true but out of reach of current technology. – Qiaochu Yuan Dec 14 '11 at 16:46
Indeed, but I thought that this very particular case might be accessible. – Alex Dec 14 '11 at 16:49
The logs of the primes are linearly independent over the algebraic closure of the rationals, by Baker's theorem. In particular, this solves the case $n=2$ of your problem and variants involving only two primes. I am not sure that helps for the general case, though. Why do you want to know? Maybe a different result will be enough for you. – Felipe Voloch Dec 14 '11 at 18:19