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When the runtime of the GNFS is given as e^(64/9*b(log b)^2)^1/3, what base is the log? I'm assuming its e, but other options would obviously be 10 and 2.

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Wikipedia gives the running time as $$ \approx \exp \sqrt[3]{\frac{64}{9}} (\ln n)^{1/3} (\ln \ln n)^{2/3}, $$ where $n$ is the integer being factored. Here $\ln n$ is the natural logarithm (logarithm to base $e$).

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