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 Additional super exponential condition andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/14/2008, 08:11 PM bo198214 Wrote:For a super logarithm the rule would be: $\text{slog}(x^y) \le \text{slog}(x) + \text{slog}(y)$ This is certainly consistent. For example: $\text{slog}(e^y) \le \text{slog}(e) + \text{slog}(y)$ $\text{slog}(y) + 1 \le \text{slog}(e) + \text{slog}(y)$ $1 \le \text{slog}(e)$ $1 \le 1$ which is true. Andrew Robbins « Next Oldest | Next Newest »

 Messages In This Thread Additional super exponential condition - by bo198214 - 10/13/2008, 07:15 PM RE: Additional super exponential condition - by andydude - 10/14/2008, 08:11 PM RE: Additional super exponential condition - by martin - 10/21/2008, 03:40 PM

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