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Invoke and/or ATTN: functional/algebraic UFies | by ihope127 | 2006-04-12 13:00:52 |
| Solve for X. |
by williamashbless |
2006-04-12 13:17:06 |
log(y) = log(x^y)
log(y) = y log x
log(x) = log(y)/y
x = 10^(log(y)/y)
as long as y is greater than 0 in this equation, x can be anything, and there will be a corresponding y. (y=0 is expressly forbidden). That restriction is also true in the original equation, so we've identified the only valid restriction.
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[ Reply ] |
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Meh, I did this backwards, didn't I. Sorry. (n/t) | by williamashbless | 2006-04-12 13:18:10 |
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Meh. On reflection, I'm hanging up my logic spurs | by williamashbless | 2006-04-12 13:20:38 |
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but 10^(log(y)/y) is convergent, no? (n/t) | by imperito | 2006-04-12 13:21:19 |
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Well, I can't solve for y directly BUT... | by williamashbless | 2006-04-12 13:29:11 |
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like I said, max of xth root of x is at e (n/t) | by imperito | 2006-04-12 13:34:01 |
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Not that I disagree, but how can you express it... | by williamashbless | 2006-04-12 13:38:43 |
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start with the original eqn... | by imperito | 2006-04-12 13:48:47 |
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Glargh, where derivative of x=f(y) is zero. | by williamashbless | 2006-04-12 14:02:29 |
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Steiner's Problem. | by imperito | 2006-04-12 14:03:54 |
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Its everywhere! Oh my gods, e is taking over! (n/t) | by williamashbless | 2006-04-12 14:09:00 |
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It feels that way sometimes | by imperito | 2006-04-12 14:13:32 |
