Timeline for Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
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| Feb 1, 2023 at 20:48 | audit | First questions | |||
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| Feb 1, 2023 at 17:22 | audit | First questions | |||
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| Jan 31, 2023 at 21:05 | audit | First questions | |||
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| Jan 31, 2023 at 14:23 | audit | First questions | |||
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| Jan 26, 2023 at 21:30 | audit | Reopen votes | |||
| Jan 26, 2023 at 22:34 | |||||
| Jan 25, 2023 at 12:00 | history | tweeted | twitter.com/StackMath/status/1618217152527454215 | ||
| Jan 15, 2023 at 17:31 | audit | First questions | |||
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| Jan 15, 2023 at 11:25 | audit | First questions | |||
| Jan 15, 2023 at 11:32 | |||||
| Jan 10, 2023 at 21:24 | answer | added | albert chan | timeline score: 2 | |
| Jan 5, 2023 at 19:58 | history | edited | amWhy | CC BY-SA 4.0 |
deleted 28 characters in body
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| Jan 4, 2023 at 16:49 | answer | added | Bob Dobbs | timeline score: 2 | |
| Jan 4, 2023 at 14:56 | answer | added | Adam Rubinson | timeline score: 4 | |
| Jan 4, 2023 at 14:39 | comment | added | Ekaveera Gouribhatla | @lulu What you said is correct, with simple manipulation we can use calculus as shown by two authors below :) | |
| Jan 4, 2023 at 14:26 | answer | added | River Li | timeline score: 18 | |
| Jan 4, 2023 at 14:15 | answer | added | achille hui | timeline score: 12 | |
| Jan 4, 2023 at 13:29 | vote | accept | Ekaveera Gouribhatla | ||
| Jan 4, 2023 at 12:56 | comment | added | Milten | If we use $\sqrt2<\frac{17}{12}$, then it's enough to show $3^{17}<2^{27}$. | |
| Jan 4, 2023 at 12:53 | answer | added | Henry | timeline score: 12 | |
| Jan 4, 2023 at 12:41 | comment | added | Arthur | Using a calculator to cheat a little, we see that $9^{\sqrt2} \approx 22.36$ while $\sqrt{2}^9\approx 22.63$. So it is, for instance, not possible to solve this by wedging an integer between them. If you square them you get $500.04$ versus $512$. Maybe those are easier to compare? | |
| Jan 4, 2023 at 12:38 | comment | added | lulu | Absolutely right, I have deleted that comment. | |
| Jan 4, 2023 at 12:38 | comment | added | Ekaveera Gouribhatla | @lulu true but wont that work only for both quantities grater than $e$ or both less than $e$. Here $9>e, \sqrt{2}<e$. | |
| Jan 4, 2023 at 12:38 | comment | added | Henry | @lulu as in another question. But here $1<\sqrt{2}<e<9$ so it is not so easy | |
| Jan 4, 2023 at 12:36 | comment | added | Ekaveera Gouribhatla | @Arnaldo yes, so i got stuck | |
| Jan 4, 2023 at 12:36 | history | edited | Ekaveera Gouribhatla | CC BY-SA 4.0 |
added 108 characters in body
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| Jan 4, 2023 at 12:34 | comment | added | Ekaveera Gouribhatla | sure thanks for your suggestion | |
| Jan 4, 2023 at 12:33 | comment | added | John Douma | Then you should add that to your proof. | |
| Jan 4, 2023 at 12:32 | comment | added | Ekaveera Gouribhatla | @JohnDouma its very simple $512<729 \Rightarrow 2^9<27^2$ | |
| Jan 4, 2023 at 12:30 | comment | added | John Douma | How do you know that $2^{4.5}\lt 27$? | |
| Jan 4, 2023 at 12:28 | history | asked | Ekaveera Gouribhatla | CC BY-SA 4.0 |