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Here is an answer that is similar in spirit to Frank and Peter's answers, but possibly simpler.

Summing by parts, we see that

$$ \sum_{p\le x} \frac{\log^k p}{p} = (\log x)^{k-1} \sum_{p\leq x} \frac{\log p}{p} -(k-1)\int_{2^-}^x (\log u)^{k-2}\sum_{p\le u} \frac{\log^k frac{\log p}{p} \frac{du}{u}.$$

Now use the formula $$ \sum_{p\leq x} \frac{\log p}{p} = \log x + c_1 + O(\exp(-c_2\sqrt{\log x}))$$ and it is not hard to derive that $$ \sum_{p\le x} \frac{\log^k p}{p} = \frac{\log^k x}{k} + c_3 + O(\exp(-c_4\sqrt{\log x})).$$

This seems easier than dealing with $Li(x)$.
    Post Undeleted by Micah Milinovich
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Similar

Here is an answer that is similar in spirit to Frank and Peter's answers, here is how to get the leading order asymptotic without using the Prime Number Theorem (in the spirit of Mertens)but possibly simpler.

Summing by parts, we see that

$$ \sum_{p\le x} \frac{\log^k p}{p} = \int_{2^-}^x (\log u)^{k-1} d\left(\sum_{p\leq u} \frac{\log p}{p} \right).$$

An integration by parts yields

$$ \sum_{p\le x} \frac{\log^k p}{p} = (\log x)^{k-1} \sum_{p\leq x} \frac{\log p}{p} -(k-1)\int_{2^-}^x (\log u)^{k-2}\sum_{p\le u} \frac{\log^k p}{p} \frac{du}{u}.$$

Now use Mertens' the formula $$ \sum_{p\leq x} \frac{\log p}{p} = \log x + O(1)$$ c_1 + O(\exp(-c_2\sqrt{\log x}))$$ and it is not hard to derive that $$ \sum_{p\le x} \frac{\log^k p}{p} = \frac{\log^k x}{k} + O((\log x)^{k-1}).$$c_3 + O(\exp(-c_4\sqrt{\log x})).$$

This seems easier than dealing with $Li(x)$.
    Post Deleted by Micah Milinovich
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