Free42 with IEEE 7542008 decimal floatingpoint  interested in a sneak preview?

03182014, 07:57 AM
Post: #87




RE: Free42 with IEEE 7542008 decimal floatingpoint  interested in a sneak preview?
(03182014 07:16 AM)Werner Wrote:(03172014 01:29 PM)Thomas Okken Wrote: 10^{n} ≤ x < 10^{n+1} → n ≤ LOG(x) < n+1This is not possible. I tend to agree with Werner, and the condition above is not needed. Strictly speaking, all we need is that 10^x is exact for x an integer, and it will be nice that LOG(x) is an integer for x exactly a power of ten. JF 

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