You Should Learn This Trick!

[@BrainStationAdvanced]

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[@xdkristof]

If you have an equation of the form a^b and b^a and you're supposed to find the bigger one, if both numbers are bigger than Euler's number (e), then the one with the bigger exponent is always the bigger number.

[@Wjneidnckrkzyk]

Thank you for this!

[@bigchungus8147]

Here is a much quicker trick (that only works if the question talks about a^b vs b^a, not when there are more than two different numbers like a^b vs c^d). If both of the numbers (an and b) are bigger than e (≈2,7182818459045) the one with the larger exponent will be larger. Like with the example with 222 and 333 the larger exponent (222^333) is larger, but compare like 2^1.5 and 1.5^2 and since both numbers are smaller than e the smaller exponent will be larger one 2^1.5 > 1.5^2

[@tai9404]

For this specific example you oculd also just do it logically like so: because 222 is 2/3rds of 333 you could just see for every 333*333 you would be able to match 222*222*222 and the latter is obviously larger

[@ravipramod1595]

Bro u should start creating educational videos on math topics such as calculus. U solve university level math problems and u have a potential in teaching. Plus u have better communication skills than I have because I freeze whenever I'm in front of a camera. 😂 Your math knowledge shall make u reach great heights 1 day❤🤟 good luck

[@BrainStationAdvanced]

Thank you so much 🙏 i will try to upload more educational videos in the future

[@Asad-g9b]

My calculator doesn't show any infinity or undefined it converts the value into power of 10 like it shows 222^333 as 2.165411765E781 and 333^222 as 9.607708951E559

[@BrainStationAdvanced]

Is it? I tried 5-6 online calculator but all of them showed error or infinite or undefined. Thanks for sharing this

[@_sIash]

my calculator didn't work so i had to make a C program, and it said that 222^333 is greater

[@TheOneAnd178]

PxEn= Px 10^n

[@Asad-g9b]

@@BrainStationAdvanced it was a built in calculator in my smartphone

[@DanMusceac]

Another way by using logaritms: ln 222^333=333.ln222. 333^222=222.ln333. Because the liniar functions grows quicker than the logaritmic function 222^333 is greater than 333^222.

[@roland3et]

Here's the trick: a^b > b^a if b>a>2 (a, b integers) Knowing this, you can decide these 'power-flippers' at your fingertips 😉. In this case 333 > 222 (> 2) therefore 222^333 > 333^222 It's as simple as that. 🙂👻

[@True_ModernUser]

Such an easy question to incremental grinders 👍

[@mkadx2136]

1) Take the root 111 from each side => 222^3 vs 333^2 or 111^3 • 2^3 vs 111^2 • 3^2 2) Divide both sides by 111^2 => 111 • 2^3 vs 3^2 => 888 > 9

[@monroeclewis1973]

Using exponents and simplifying you eventually get down to (2^111 x 2^111 x 2^111 x 111^111) vs (3^111 x 3^111). Now take both sides to the 1/111th power. You are left with (2 x 2 x 2 x 111) vs (3 x 3), or 888 vs 9. Because 888 > 9, the first expression is greater. No need for logs.

[@goggypoggy]

did you watch the video???

[@wowowo3027]

X^Y < Y^X when 0=

[@ShanthiVeluswamy]

No, 2^3 is smaller than 3^2

[@Shelzastark]

Can we use logarithms here??

[@BrainStationAdvanced]

Pls check the end part of the video...also solved this using logarithm

[@RyanLewis-Johnson-wq6xs]

222^333>333^222 It’s in my head.

[@AarushDutta-v9u]

Same bro, it was that simple. The exponent 333 is so much more than 222

[@pratyushkhandelwal7138]

Thumbnail is so funny 😂😂

[@BrainStationAdvanced]

😗

[@pankusharma1971]

According to me 333^2 < 222^3 is great so 333^222 < 222^333 😅

[@lukatolstov5598]

I found out, that 2³>3², but 2⁴=4², but also 3⁴>4³. The greater number becomes, the bigger X^N where X

[@Mr_Deuch_guy]

ohh i get it

[@SuperCreepo1809]

Please Pin

[@BrainStationAdvanced]

Sorry but that is already reserved

[@SuperCreepo1809]

@BrainStationAdvanced Ok😥🥴🥴

[@bikrampratapbhatta9797]

More video need on Calculus

[@BrainStationAdvanced]

Yeah waiting for the likes criteria to be fulfilled.