| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1501.1 | x= log(10-10^y) | STAR::ABBASI | | Tue Oct 15 1991 11:10 | 1 |
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| 1501.2 | | CLT::TRACE::GILBERT | Ownership Obligates | Tue Oct 15 1991 21:48 | 1 |
| re .1: That answers a similar question. Close, but no cigar :^).
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| 1501.3 | y = 1 + log ( 1 - 10^(x-1) ) | QUIVER::COUCOUVITIS | | Wed Oct 16 1991 08:44 | 2 |
| is that closer ?
arthur
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| 1501.4 | nope | ALLVAX::JROTH | I know he moves along the piers | Wed Oct 16 1991 09:17 | 3 |
| Re .3
- Jim
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| 1501.5 | | WONDER::COYLE | | Wed Oct 16 1991 10:11 | 5 |
| How about:
Y = log( 10 - 10^X )
-Joe
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| 1501.6 | | GUESS::DERAMO | Dan D'Eramo, zfc::deramo | Wed Oct 16 1991 10:19 | 1 |
| :-)
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| 1501.7 | | WONDER::COYLE | | Thu Oct 17 1991 10:23 | 14 |
| I beleive if you relook at .3 you will see that it is a diferent form
of .5:
Reply 5 >> Y = log( 10 - 10^X )
Y = log( (10) * (1 - 10^(X-1) )
Y = log( 10 ) + log( 1 - 10^(X-1) )
Y = 1 + log( 1 - 10^(X-1) )
Which is the same as Reply 3
-Joe
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| 1501.8 | f <==> g | STAR::ABBASI | | Sat Oct 19 1991 17:24 | 3 |
| the question of if forms of expression are equivlant is very important
in computer algebra. since then one chooses the form that has less
complicated internal structure.
|