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Additional super exponential condition

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Additional super exponential condition
andydude Offline
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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
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RE: Additional super exponential condition - by andydude - 10/14/2008, 08:11 PM

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