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If an algorithm for $SAT$ runs in $O(n^{\log n})$ time, and if $L$ belongs to $\mathsf{NP}$, is there an algorithm for $L$ that runs in $O(n^{\log n})$ time?

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As $\mathrm{SAT}$ is $\mathrm{NP}$-complete, we know that there is a polytime reduction from our language $L$ to $\mathrm{SAT}$. However, this reduction can involve a polynomial blowup of the input size. This means that an input $w$ to $L$ could be mapped to a formula $\phi_w$ such that $|\phi_w| \approx |w|^k$ for some constant $k$.

Running our hypothetical algorithm for $\mathrm{SAT}$ on $\phi_w$ takes time $O(|\phi_w|^{\log |\phi_w|}) = O((|w|^k)^{\log |w|^k}) = O(|w|^{k^2 \log |w|})$ time. But $O(n^{k^2 \log n}) \neq O(n^{\log n})$.

Thus, there is no immediate reason why the assumption would yield an $O(n^{\log n})$-algorithm for $L$.

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Since any input x for L can be reduced to SAT in O(|x|^c) time, the created SAT instance will have size at most n=O(|x|^c). So the n^(lg n)-time algorithm for SAT will be an |x|^O(lg |x|)-time algorithm for L.

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