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December 3

If a^b = a^c, then b = c.

In the article section: Mathematical_fallacy#Proof_that_x_.3D_y_for_any_real_numbers_x_and_y it says:

"The error in this proof lies in the fact that the stated rule is true only for a positive number a which is not equal to 1."

It seems to me however, that actually the number a must not equal to ±1 or 0, and that it does not necessarily need to be positive.

Am I right? (Also, does the number need to be real? Or rather is it, that the real portion of a needs to follow the rule above?) Ariel. (talk) 11:23, 3 December 2009 (UTC)[reply]

As long as b and c are integers then I think this "rule" is true for any non-zero real or complex value of a provided that a is not a root of unity. For example, the rule does not hold for a = (1+i)/√2 because a⁸ⁿ⁺¹ = a for any integer n. If b or c are not integers then a^b and/or a^c are not single valued, so you have to be specific about which of their values you are using. Gandalf61 (talk) 12:02, 3 December 2009 (UTC)[reply]

If i⁴ = i⁸, then 4 = 8. I rest my case. 139.130.57.34

the number a must not equal to ±1. CHECK
the number a must not equal to 0. CHECK
the number a does not necessarily need to be positive. CHECK

139.130.57.34 (talk) 21:05, 3 December 2009 (UTC)[reply]

Your case fails: Gandalf said "provided that a is not a root of unity". Obviously i is a root of unity. Michael Hardy (talk) 01:39, 4 December 2009 (UTC)[reply]

It seems 139's case was simply to refute the OP's conjecture - in a simpler way than Gandalf's - and in that he succeeded. -- Meni Rosenfeld (talk) 08:25, 4 December 2009 (UTC)[reply]

OK, I just looked at the article that was linked to. It does say that b and c must be real. It assumes at most tacitly that a is real. Michael Hardy (talk) 22:53, 4 December 2009 (UTC)[reply]

There is an issue of whether b and c are required to be real. If one allows complex b and c, then here's a counterexample: e^2πi = e⁰. Michael Hardy (talk) 01:41, 4 December 2009 (UTC)[reply]

I updated the section in the article, I hope it's accurate. Ariel. (talk) 09:42, 7 December 2009 (UTC)[reply]

Growth of Log and Power

I'm trying to find an answer to a problem that I'm sure is very easy, but I am way too tired to wrap my brain around it and searching isn't turning up the topic I want... I understand well that n² grows faster than n and log n grows slower than n. Is there a specific number x such that log^x n grows linearly just like n? Basically, what power of log cancels out the growth of both the power and the log? My assumption is that it is e, but that is not how e is defined. -- kainaw ™ 12:44, 3 December 2009 (UTC)[reply]

log^x n grows slower than n for every constant x. You'd need x = (log n)/(log log n). — Emil J. 12:47, 3 December 2009 (UTC)[reply]

Thanks. Now that I'm a little more awake I can see it much clearer. -- kainaw ™ 13:28, 3 December 2009 (UTC)[reply]

December 3

If ab = ac, then b = c.

Growth of Log and Power

If a^b = a^c, then b = c.