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I have come up with a way to compare large exponents, for example: I can tell which number is bigger in $12345^{78901}$ or $21346^{78900}$ within a few seconds without using calculator. So I have 2 questions.

1)Is it something important?(I dont know if it something simple that anyone can do) Can you compare them easily?

2)If it is something useful how and where can I publish it?

(I didn't post it on academia stack exchange because of the first question)

Update:

$9873^{64}$ > $11424.9^{63}$(took me about 5 seconds to compare , using only pen and paper with no calculator)

I don't know if this was a good example but I can do the same for even bigger numbers in which the answers are very close to each other. Seeing the comments I think it is a pretty good method, so how do I let people in masses to know about it. It is not a research paper but just a simple method so, is there any magazine that publishes this kind of stuff?

Pranav K
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  • It depends on what the technique is. – Randall Jan 28 '20 at 16:02
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    Without seeing the technique in question there's really nothing to be said. Note: that example is not a compelling one, since the second expression is $\textit {a lot} $ greater than the first. – lulu Jan 28 '20 at 16:05
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    Try this one: $10375^{105}$ versus $70288^{87}$. The ratio (in one direction or the other) is about $1.00020$ so it's hard to imagine that any crude estimate is going to settle the point. – lulu Jan 28 '20 at 16:16
  • I like that example, @lulu , because $105 \lfloor \log_2(10375) \rfloor>87 \lfloor \log_2(70288) \rfloor$ and the same for the ceilings, but the inequalities without rounding go the other way. – Ian Jan 28 '20 at 16:29
  • @Ian Right, this one seems tough. I'd be impressed by a universally applicable pencil and paper method that quickly resolved it. – lulu Jan 28 '20 at 16:31
  • @lulu that one reduces to $10375^{35}$ and $70288^{29}$ which then has a ratio of :$$1.000067460485559641733994215\ldots$$ –  Jan 28 '20 at 17:30

2 Answers2

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You can break this problem down using some pretty gross estimations.

e.g. $21346 > 1.5\cdot 12345 > \sqrt 2\cdot 12345$

$21346^{78900} > \sqrt 2^{78900} \cdot 12345^{78900}$

$12345^{78901} = 12345\cdot 12345^{78900}$

If $\sqrt 2^{78900} > 12345$ we are done.

$2^{20} \approx 1$ million

$\sqrt 2^{40} > 12345$

$21346^{78900}$ is significantly bigger than $12345^{78901}$

Doug M
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  • Hey! I realize that its a bad example but my method doesn't rely upon estimation, even if the 2 numbers are very big and give results that are very close I can still calculate them easily with just pen and paper. Anyway, my question was if there is any magazine that publishes such methods to solve textbook problems. Thank you for the help. – Pranav K Jan 28 '20 at 17:27
  • @PranavK it'll depend on what math it uses ... –  Jan 28 '20 at 17:35
  • @RoddyMacPhee, highschool level math. – Pranav K Jan 28 '20 at 17:36
  • You can memorize $\log 2 \approx 0.3$ and with this information, you work out a decent fraction on the log table. With these numbers, $21346 > 2\times 10^4$ and $12345 < 2^\frac {-3} \times 10^6, \log 21346 > 4.3, \log 12345 < 4.1$ – Doug M Jan 28 '20 at 21:35
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The method you proposed in your (now deleted) post is not valid. It is simply not true that $$a>b\implies x^a>y^b \quad \forall x, y$$

Specifically, you err when you assert that $$\frac {\log_c A}{\log_c B}=\log_c (A-B)$$

To see the difficulty of the problem, note that, while it is true that $$10375^{105}> 70288^{87}$$ we also have $$10375^{105}< 70289^{87}$$

Thus, whatever method you use it must be sensitive enough to detect the difference between $70288$ and $70289$. That's why I suggested that this would be a difficult test case.

lulu
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