Comparing {sqrt(2)}^{sqrt(7)} and {sqrt(3)}^{sqrt(3)}

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Комментарии
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You did put the cart in front of the horse - means you knew the answer and stared from 2^14 against 3^9.
Normally you had to go other way. and come to 2^7 against 3^√21. Then logically you start bounding √21. The simplest guess is that √21>4.5 what is fortunately true. Then to get rid of 0.5 you square both sides/ Same what you did but in a natural way.

vladimirkaplun
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i think you need to review the part from 2:30 to 2:50 !

abdeba
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You can also directly calculate the numbers in excel and compare. The story is valid because it concerns positive numbers and squaring c.q. square rooting (being increasing functions then) does not change the order.

mystychief
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I would guess {sqrt(k)}^{sqrt(k)} is always > {sqrt(k-l)}^{sqrt(k+l)} as long as k-l>0.

JXSJ
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Elevo al quadrato.. 2^sqrt7 or 3^sqrt3.. Elevo a sqrt17, risulta 2^sqrt119 or 3^sqrt51..il primo numero è <2^11=2048..il secondo è >3^7=2187...

giuseppemalaguti
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I can't say that my approach to solving this was valid, but I still guessed correctly!

scottleung
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So thought_provoking...
Good job...👌👌👌.

mehrdadbasiri
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SyberMath, I believe I've solved your problem in a much simpler way! Can I send you my solution?

syedmohammadhasanbintariqu
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Jake multiple √3 by ln√3/ln√2 and see if it is greater than √7 and u are done

Chris_
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