Let's Compare Two Numbers

[@-basicmaths862]

17^20>16^20, 64^13>63^13 17^20>4^40, > 4^39>63^13

[@bjorntorlarsson]

Of course! One should perhaps not go down to base 2 immediately, but just halve it and look for an obvious answer.

[@pwmiles56]

17"20 / 63^13 = 289^10 / 63^13 = (289/63)^10 / 63^3 > 4^10 / 63^3 = 4x64^3 / 63^3 > 4 > 1

[@TedHopp]

So of the two methods listed at the start of the video, in the end neither was used. Instead, a b was decided indirectly, using something like a > c d > b. Maybe we could call it the method of surrogates. Of course, if one finds that c < d, that approach fails to draw a conclusion one way or the other. One then has to look for different surrogates or try a different method.

[@bjorntorlarsson]

I happened to find my quite old calculator the other day. So I put 17^20 into it, and it said that it is 3. At least if one turns it upside down, otherwise it is E, which would be 14 in hexadecimal. And hex is beyond its specifications. So we do need another method. (It kinda looks like a little pony if one turns it 90 degrees. Like a dead one the other way.)

[@SyberMath]

😁

[@Kounomura]

Predicate: 17^20 > *16^20 > 64^13* > 63^13 ---> 16^20 > (4*16)^13 ---> 16^7 > 4^13 --> 4^14 > 4^13 qed.

[@forcelifeforce]

*@ SyberMath* 17^20 > 16^20 = (4^2)^20 = 4^40 > 4^39 = (4^3)^13 = 64^13 > 63^13 Therefore, 17^20 > 63^13.

[@hazalouldi7130]

tu as commis une erreur à5min pour la forme binomiale,on devrait avoir >1.Merci

[@rob876]

17^20 > 16^20 = 2^80 63^13 < 64^13 = 2^78 63^13 < 2^78 < 2^80 < 17*20

[@nathanenzo6807]

Nice

[@SyberMath]

Thanks

[@receparikan639]

Always bet on the bigger exponent (:

[@RealQinnMalloryu4]

17^2^10or6^1^1^1 17^12^10or6 1^12^2^5or2^3 (x ➖ 3x+2).1^2^1or 2^1 (x ➖ 2x+1). 2^1>2^3 17^20>63^13

[@Lilobababoi8294]

idk hahhaa my head dizzy

[@radupopescu9977]

I was sure that you would use logarithms... i was wrong, but it would be easier!

[@SyberMath]

Logs can’t be used! That would be using like using a calculator

[@radupopescu9977]

@@SyberMath Not really. There were books for logs when I was younger.... you were allowed to use those books (tables with logs values).

[@scottleung9587]

I used the powers of 2 to solve this - couldn't believe how easy it was!

[@SyberMath]

You are good!

[@giuseppemalaguti435]

17^13*17^7 or 63^13..17^7 or (63/17)^13...ma si ha che (63/17)^13(63/17)^13..quindi risulta 17^20>63^13....mah,non sono sicurissimo del ragionamento

[@wjrasmussen666]

The only thing you can use is your brain. me: crap.

[@SyberMath]

😁