I think the answer to this problem is $1012$, but I'm not sure.
Any ideas? Any and all help is appreciated.
The answer is quite straight forward: $S=\mathrm{Re} P(1)=2^{1012}$, so $\log_2 S=1012$.
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The answer is quite straight forward: $S=\mathrm{Re} P(1)=2^{1012}$, so $\log_2 S=1012$.