I instantly chose the first expression simply because the exponent is always the important part of such an expression when evaluating. Especially with large numbers.
@@jeffrybassett7374 you still need to do a quick evaluation. For instance 33^33 is less than 88^22 but I think the quick trick of treating them as 8^2 vs 3^3 should get you right answer on which is larger.
Given the nature of exponents, I just sortof figured that such a bigger exponent would be a bigger number, but I really enjoyed your explanation and proof! Thanks for putting it together and showing us the steps.
yeah i did this mentally, simply based on the fact that for such large exponents its really always going to be the lower number to the higher power thats bugger
Nice! Here’s what I did: If we can show that 222^333 / 333^222 >1, then 222^333 is larger. I factored out 222^111 to get (222/333)^222 * 222^111. The fraction in the parentheses reduces to 2/3. If we factor a 2 out of the exponent and distribute it within the parentheses, we get (4/9)^111 * 222^111. Now that they both share the same exponent, we can write them together as (4/9 * 222)^111. Multiplying in the parentheses, we get 888/9, a number just under 100, which is certainly greater than 1. Raising this number to the power of 111 will only make it bigger. Thus, because the fraction was ultimately >1, 222^333 is larger than 333^222.
Use log as it is known as increase function for x>0 plane. log(222^333) = 333(log2+log111) = 333(0.301+log111)--(a) log(333^222) = 222(log3+log111)= 222(0.47+log111)--(b) (a)/(b) = (3/2)(0.301+log111)/(0.47+log111) >1 therefore a>b and it means 222^333>333^222
I just want to add that showing all the steps regardless if we should know it or not, is something I've always thought is a sign of a good teacher. Keep those videos coming!!!
222^333=(222) ^3) ^111 333^222=(333) ^2) ^111 Now, you can easily see which one is greater. If you can't see that then simply divide one from another. Here, in this case 222*222*222/333*333=4*222/9
I’m not good at math but I could spot that answer immediately. The difference between the base numbers is small. The power 333 is obviously vastly larger than the power of 222
if we notice the expression: f(x) = (x^(1/x222*333)) where x = 222 and 333 we can use the concept of increasing decreasing functions and find the critical point which comes out to be e. since 222 and 333 both are greater than e then x^1/x is decreasing in x for x>e hence correct answer will be 222^333
Nice method. I went about it a different way, a bit quicker, creating equivalent exponents ... 222^333 = 333^222 111^333 . 2^333 = 111^222 . 3^222 111^222 . 111^111 . 2^333 = 111^222 . 3^222 Now, 111^222 is a common factor of both sides, so we really just have to show that 111^111 . 2^333 is = 3^222 I did so by changing the base of 3^222 to base 9 ... (I have written "9 to the half" as 9^0.5) 111^222 . 111^111 . 2^333 = 111^222 . (9^0.5) ^222 111^222 . 111^111 . 2^333 = 111^222 . 9 ^111 clearly 111^111 alone is > 9^111 so 111^222 . 111^111 . 2^333 > 111^222 . 9 ^111 and 222^333 > 333^222 bit ugly typing on the computer., but looks neat and elegant on paper :(
My way is :- 222^333 = (2 x 111)^(111 x 3) & 333^222 = (3 x 111)^(111 x 2) So you can take the 111th root of each side and then you only need to establish if (2 x 111)^3 is larger then (3 x 111)^3, which is the same as 2^3 x 111^3 and 3^2 x 111^3. Divide both by 111^2 and you get 2^3 x 111 and 3^2, which is 888 and 9 It's clear therefore that 222^333 must be greater than 333^2222
If you have a calculator, obviously punching the equation in won’t work but you can simply use e^ln(a^b) = e^(b.ln(a)). You end up with 333.ln(222) and 222.ln(333) which a calculator can handle. Can use this trick for all of these kinds of problems
I knew the answer was the first one because although the initial base (3) was greater than (2), the power 333 is far greater than 222. It's just like comparing 2⁵⁰ to 5²⁰, where 2⁵⁰ is greater because it's the same as 32¹⁰ while 5²⁰ is same as 25¹⁰.
We can simply take log to base 111 on both sides. Then ignore log to base 2 and 3 in sum of logarithms as they will be close to zero and much smaller than one.
Took 5 minutes but I worked out on my own. :D 222^333 = 37^333 x 2^333 x 3 ^333 333^222 = 37^222 x 3^444 Divide both sides for ((37 ^222) x (3^333)) and you get 37^111 x 2^333 and 3^111 Since 37^111 > 3^111, 37^111 x 2^333 > 3^111 In conclusion, 222^333 > 333^222
I think it is easier to realize that 222^333= (222^3)^111 and 333^222=(333^2)^111. Then we only need to compare 222^3 and 333^2, you dont even need a calculator to determine that 222^3 is bigger due to digits
Rewriting 333 as 222x1.5 and working from there makes the problem easier. The common factor of 222^222 can be removed from each with 1.5^222 compared to 222^111. 1.5^222 is (9/4)^111 which is less than 222^111
1) 222³³³ = 222²²² x 222¹¹¹ 2) 333²²² = (222 x 1.5)²²² = 222²²² x 1.5²²² 3) Devide both numbers to 222²²² and compare the rest: 222¹¹¹ or 1.5²²² 4) 1.5²²² = (1.5²)¹¹¹ = 2.25¹¹¹ 5) Answer 222 > 2.25 so 222³³³ > 333²²² p.s. I dont even remember those rules from the school, just imagine an array of multiplying numbers and what you can do with them. For example if you have an array of 1.5 take every next two and multiply them, youll get an array of 2.25 but it is two times shorter.
The power component dominates the size of the number. If you take a log, 333 times very small number vs 222 times very small number. So obviously the former is larger than the latter.
Intuitively it has to be the first one. Proof? 222^333=(222^1,5)^222=(222*√222)^222 Now because √222>√100=10 we can conclude that 222*√222>2220>333. From this it follows that (222*√222)^222>333^222 and hence by the equation I started with that 222^333>333^222 qed.
Though what you say is true, I think this kind of problem should typically not be seen as you say, because in such problems the "n" are not any n and especially not the same, they are choosen in a specific way in order to be related with one another... and there's obviously no reason to have for any function f : O(n^333) >> O(f(n)^222) In particulzr for 2^333 and 3^222 this does not work anymore
The other - but very simillar - solution is: On left side we have 111 mutiplications of 222^3 on the right we have 111 mutiplications of 333^2. Left side is 111^2*111*8 on right side we have 111^2*9. Since 111 mutiplications of 888 > 111 mutiplications of 9 then left side is bigger:)
"Ou" é o mesmo que "menor maior ou igual", ou seja, um sinal matemático. Tudo q eu fizer de um lado, tomando esse raciocínio, posso fazer do outro. Oq eu fiz. 222³³³ or 333²²² posso escrever como 222³×¹¹¹ or 333²×¹¹¹. Tirando raiz 111 dos dois lados, cancela, então 111 some dos expoentes😮. Depois disso, posso deixar a minha ignorância, e tentar tambem escrever 222 e 333 como, (2×111)³ or (3×111)². Meu celular vai descarregar, e n dá pra falar tudo, mas desenvolve isso aí q vai dar certo
Have not yet watched the video, but: say X = 222. Then 1.5x=333 You are now comparing x^333 vs (1.5x)^222. Distributing the exponent 222, you get... X^333 vs (x^222)(1.5^222). Divide both sides by x^222... X^111 vs (1.5) ^222. The right hand term, however, can be expressed as 1.5^(2*111), which can in turn be expressed as (1.5^2)^111. X^111 vs (1.5^2)^111. Raise each side to the exponent (1/111) to get rid of the ^111 part... X vs (1.5)^2. Since X is 222, we have 222 vs 2.25. 222>2.25, and thus we arrive at: 222^333 > 333^222.
maths just say if the comparison of two numbers, and their sum of cubic or square root is greater then the outcome number is greater, so start from smaller square, and smaller cube, so , if 222 cube is smaller than 333 squared, the the greater number will be always greater we obtain 10 941 048 is greater than 110 889
My head made both exponents to 666 like this: 222^333 or 333^222 sqrt2(222)^666 or sqrt3(333)^666 Then I cancel both exponents and have sqrt2(222) or sqrt3(333). I know the first is just smaller than 15, so take 14. And then check the right side where 14*14*14 is much larger than 333, so the sqrt3(333) is therefore smaller than sqrt2(222). Thus 222^333>333^222
So my guess is that for all numbers larger than 1: x^y > y^x if y is bigger than x e.g. 5^6 > 6^5 And for all numbers smaller than 1: x^y < y^x if y is bigger than x e.g. 0.5^0.6 < 0.6^0.5
So much math needed over a simple question wich is just the same as: Wat is bigger? 2x5 or 5x2... The only difference is that you use bigger numbers in this problem but the answer is the same. So you might wanna redo your math here. After all: 222*333 = We dont know (111*2 = 222)*333 = Same as line 1 111*333 = (Donno exact number)e+370 = Not even close to being 222? + 2*333 < Also 2x more then on line 2? So yeah even without calculator i can tell your math is of. Because on line 2: 2*333 = (with calculator = 3,533694129556769e+72) < i would not know However on line 3: (2*3*111 = (2x2x2 = 8))x111 = 888? < would know this Conclusion the lines you write out make no sense in the language of Mathematical algebra. Even in language of hex code that would be 888136 888136 888136 over code 332193 332193 332193. Wich roughly translates to 888 and 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000? Wich makes even less sense to even think they would be near the same. Conclusion do not add more lines of math to make it more easy, As it will only make it more easy to make mistakes. I hope you take this as constructive feedback rather then just a comment that tells you, That you are wrong. Because it is not my intention to declare anyone of mistakes. I only want to point out more learning opportunities for everyone to come to the right conclusion. (Even if that means i am wrong.) I hope this helps bring clarity and have a nice day.
*@Learncommunolizer* -- Here it is using logarithms without guesswork: 222^333 vs. 333^222 333*log(222) vs. 222*log(333) Divide each side by 111 and expand inside the parentheses: 3*log(111*2) vs. 2*log(111*3) 3*[log(111) + log(2)] vs. 2*[log(111) + log(3)] 3*log(111) + 3*log(2) vs. 2*log(111) + 2*log(3) log(111) + 3*log(2) vs. 2*log(3) log(111) + log(8) > log(9) Therefore, 222^333 > 333^222.
@@gromosawsmiay3000 It's not so hard that you need a calculator for it. 1.5^2=1.5*1.5, and multiplying a number by 1.5 is the same as adding half of the number to itself. Half of 1.5 is 0.75, so 1.5+0.75=2.25.
@@aj_style1745 I do not need calculator, sometimes only pencil and paper, but this was example how to solve this exercise without calculation. I remember my time on technical university.... I solved integrals in my memory when I travel by bus :-)
I instantly chose the first expression simply because the exponent is always the important part of such an expression when evaluating. Especially with large numbers.
Can’t you just short circuit everything once the numbers are the same and evaluate 2^3 vs 3^2 and come to the proper evaluation real quick.
@@m.hasler7263 No need to evaluate anything here. Unless you're really close to the origin the higher exponent will always win out.
Exactly, one higher exponent makes already the difference.
@@jeffrybassett7374 you still need to do a quick evaluation. For instance 33^33 is less than 88^22 but I think the quick trick of treating them as 8^2 vs 3^3 should get you right answer on which is larger.
@@m.hasler7263 33^33 is greater than 88^22
You can also use calculus to show that s^b > b^s whenever b > s > e.
Traceback (most recent call last):
File "", line 1, in
NameError: name 'e' is not defined
@@whatkidsandbabieslike7884 Well, the formulation may be a bit confusing, but with e I obviously meant exp(1).
Defined the big number to enter the queque..
@@whatkidsandbabieslike7884 11q1
You should be making the videos.
Given the nature of exponents, I just sortof figured that such a bigger exponent would be a bigger number, but I really enjoyed your explanation and proof! Thanks for putting it together and showing us the steps.
Glad you enjoyed it!
Great to hear
Wish I had a math teacher like you in my HS days!👍👍👍
Wow, thanks 👍👍👍
I agree, these are entertaining. And OCD compliant. Handwriting is excellent. Cheers!
yeah i did this mentally, simply based on the fact that for such large exponents its really always going to be the lower number to the higher power thats bugger
very logical approach to solve the question.
Excellent simple and beautiful method, thanks Professor.
You are welcome! Thanks 🙏❤️🙏
I guessed right, but your mathematic deduction was quite impressive and yet easy to follow. 👍
Well done! 👍👍👍
Nice! Here’s what I did:
If we can show that 222^333 / 333^222 >1, then 222^333 is larger. I factored out 222^111 to get (222/333)^222 * 222^111.
The fraction in the parentheses reduces to 2/3. If we factor a 2 out of the exponent and distribute it within the parentheses, we get (4/9)^111 * 222^111. Now that they both share the same exponent, we can write them together as (4/9 * 222)^111. Multiplying in the parentheses, we get 888/9, a number just under 100, which is certainly greater than 1.
Raising this number to the power of 111 will only make it bigger. Thus, because the fraction was ultimately >1, 222^333 is larger than 333^222.
Nice one bro!
We're thinking the same!
Use log as it is known as increase function for x>0 plane.
log(222^333) = 333(log2+log111)
= 333(0.301+log111)--(a)
log(333^222) = 222(log3+log111)=
222(0.47+log111)--(b)
(a)/(b) = (3/2)(0.301+log111)/(0.47+log111) >1 therefore a>b and it means 222^333>333^222
Introducing log in order to solve this problem is like using heavy equipment in just picking up a pea. It is adding more problems than easing it.
bros using logs without a calculator
@@MrSinusu bro both methods are easy enough.
@@100iqgaming yeah without calculator, of course, everyone knows how big log2 and log3 are and simply log111 is not far away from 2.
@@beckenbauer1974 what? with what base are you using, also how do you know what log 2 is? when do you do logs without a calculator outside of algebra?
I just want to add that showing all the steps regardless if we should know it or not, is something I've always thought is a sign of a good teacher. Keep those videos coming!!!
222^333=(222) ^3) ^111
333^222=(333) ^2) ^111
Now, you can easily see which one is greater. If you can't see that then simply divide one from another. Here, in this case
222*222*222/333*333=4*222/9
Tres bien
Love your videos. Even though I'm retired and no longer need to manipulate equations, these videos are really fun. Thanks
You're very welcome!
The best explain
Thanks and Welcome! 🙏❤️🙏
I guessed correctly but this way gives you the actual ratio in a nice form
beatiful decision
Nice calculation
Thanks and Welcome! 👍👍👍
Even if eyeballing, we can assume that more iterations beats a higher base almost every time unless the base is much larger
Truly nice handwriting
Many many thanks! 😀
I’m not good at math but I could spot that answer immediately. The difference between the base numbers is small. The power 333 is obviously vastly larger than the power of 222
Very nice work!!!
Thank you! Cheers!
if we notice the expression:
f(x) = (x^(1/x222*333)) where x = 222 and 333
we can use the concept of increasing decreasing functions and find the critical point which comes out to be e. since 222 and 333 both are greater than e then x^1/x is decreasing in x for x>e hence correct answer will be 222^333
Root both sides with ^(1/111) and get 222^3 vs 333^2, divide both sides by 222^2: get 222 > (333/222)^2=2.25. Done! Enjoy!
without calculation, you did calculation 333/222)^2=2.25 :-)
@@gromosawsmiay3000333/222 is 1.5
1.5² is 2.25
A nice math problem indeed, so happy to see it❤
Glad you liked it!
Thank you! 😃
Отличная задачка.
The most convoluted math problem I have ever seen.
a^b is bigger when e
amazing methods
Many many thanks !
Power of exponential
Nice
Thanks
Obviously the first one
Exactly
Nice method. I went about it a different way, a bit quicker, creating equivalent exponents ...
222^333 = 333^222
111^333 . 2^333 = 111^222 . 3^222
111^222 . 111^111 . 2^333 = 111^222 . 3^222
Now, 111^222 is a common factor of both sides, so we really just have to show that 111^111 . 2^333 is = 3^222
I did so by changing the base of 3^222 to base 9 ... (I have written "9 to the half" as 9^0.5)
111^222 . 111^111 . 2^333 = 111^222 . (9^0.5) ^222
111^222 . 111^111 . 2^333 = 111^222 . 9 ^111
clearly 111^111 alone is > 9^111 so 111^222 . 111^111 . 2^333 > 111^222 . 9 ^111 and 222^333 > 333^222
bit ugly typing on the computer., but looks neat and elegant on paper :(
Nice proof.
Indeed! Thanks 🙏❤️🙏
cool teacher
Thanks and Welcome 🙏❤️🙏
One one one (in my head)
3:11 222^333 is the better number ❤
Very nice video! Only suggestion I have is that you might have announced the reasoning behind your approach before jumping in!
Thanks for the tip!
My way is :-
222^333 = (2 x 111)^(111 x 3) & 333^222 = (3 x 111)^(111 x 2)
So you can take the 111th root of each side and then you only need to establish if (2 x 111)^3 is larger then (3 x 111)^3, which is the same as 2^3 x 111^3 and 3^2 x 111^3. Divide both by 111^2 and you get 2^3 x 111 and 3^2, which is 888 and 9
It's clear therefore that 222^333 must be greater than 333^2222
POV: you do the wrong working out but still get the correct answer
I wish I have a math teacher like u who explains soo welll very welll explained loved it
I mean if this was multiple choice, conceptually understanding exponents would clearly have 222^333 as the answer
If you have a calculator, obviously punching the equation in won’t work but you can simply use e^ln(a^b) = e^(b.ln(a)). You end up with 333.ln(222) and 222.ln(333) which a calculator can handle. Can use this trick for all of these kinds of problems
Nicely done!
Feels right to choose the first option since the beginning, but your solution is very beautiful!
Regards from Brazil!
Thank you very much!
I knew the answer was the first one because although the initial base (3) was greater than (2), the power 333 is far greater than 222. It's just like comparing 2⁵⁰ to 5²⁰, where 2⁵⁰ is greater because it's the same as 32¹⁰ while 5²⁰ is same as 25¹⁰.
Me, with a smooth brain:
2^3 < 3^2,
So 222^333 < 333^222
is likely
And boy was I wrong
We can simply take log to base 111 on both sides. Then ignore log to base 2 and 3 in sum of logarithms as they will be close to zero and much smaller than one.
Took 5 minutes but I worked out on my own. :D
222^333 = 37^333 x 2^333 x 3 ^333
333^222 = 37^222 x 3^444
Divide both sides for ((37 ^222) x (3^333)) and you get
37^111 x 2^333
and 3^111
Since 37^111 > 3^111, 37^111 x 2^333 > 3^111
In conclusion, 222^333 > 333^222
Pretty easy to proof
I think it is easier to realize that 222^333= (222^3)^111 and 333^222=(333^2)^111. Then we only need to compare 222^3 and 333^2, you dont even need a calculator to determine that 222^3 is bigger due to digits
Usually number with higher power is greater in such type of questions.
Not true. 3^2>2^3.
@@trumplostlol3007 true for both numbers > e
That's why I started with word 'usually'.
@@Dumpy332 gud btw can we solve these type of questions using binomial theorem?
@@kumarharsh837 Maybe. I've not thought about it.
Rewriting 333 as 222x1.5 and working from there makes the problem easier. The common factor of 222^222 can be removed from each with 1.5^222 compared to 222^111. 1.5^222 is (9/4)^111 which is less than 222^111
1) 222³³³ = 222²²² x 222¹¹¹
2) 333²²² = (222 x 1.5)²²² = 222²²² x 1.5²²²
3) Devide both numbers to 222²²² and compare the rest: 222¹¹¹ or 1.5²²²
4) 1.5²²² = (1.5²)¹¹¹ = 2.25¹¹¹
5) Answer 222 > 2.25 so 222³³³ > 333²²²
p.s. I dont even remember those rules from the school, just imagine an array of multiplying numbers and what you can do with them. For example if you have an array of 1.5 take every next two and multiply them, youll get an array of 2.25 but it is two times shorter.
I did this by a similar method, and I was really surprised by the result.
When comparing a^b vs b^a. If both a and b are greater than e, then a^b will always be greater
It's nice brother, 💖, 🙊🙊maths
Thanks ✌️and Welcome 🙏❤️🙏
@@learncommunolizer 💝🙈
@@MwUcyan bruh
The power component dominates the size of the number. If you take a log, 333 times very small number vs 222 times very small number. So obviously the former is larger than the latter.
Когда такая огромная разница в степенях, можно на глаз определить
Log both, and bcs log function increases slower than linear function, done
Its easy. Just simplify it. 2*2*2 is 8. 3*3 is 9 therefore 333 to the power of 222 is bigger
Hahaha. That was the most convolutedly bizarre way to do that, but okay. At least you got the right answer.
I like the part where you said "one one one" 😄
I like the part where he says two two two
I like the part where he said "eight eight eight"
alternate title:Me flexing my math skills
Usually place your bet on the smaller number to the higher power, over the higher number to smaller power. This is because of the power of powers.
Exponent is always bigger
Its easy, imagine the exponents are smaller.
2^3 vs 3^2
That's the same as 4^2 vs 3^2
Bigger exponents only exacerbate this difference.
Intuitively it has to be the first one. Proof? 222^333=(222^1,5)^222=(222*√222)^222 Now because √222>√100=10 we can conclude that 222*√222>2220>333. From this it follows that (222*√222)^222>333^222 and hence by the equation I started with that 222^333>333^222 qed.
We can think of it as time complexity, O(n^333) >>> O(n^222)
Though what you say is true, I think this kind of problem should typically not be seen as you say, because in such problems the "n" are not any n and especially not the same, they are choosen in a specific way in order to be related with one another... and there's obviously no reason to have for any function f : O(n^333) >> O(f(n)^222)
In particulzr for 2^333 and 3^222 this does not work anymore
With 111 more powers it's going to walk it!
A bigger, power of is very bigger, simply I choose when I see question it self
I know that the larger exponent will be bigger, but it's cool to see the proof
Let x=111; then, (2x)^(3x) or (3x)^(2x) ; extract x root; (2x)^3 or (3x)^2 ; 8x^3 or 9x^2 ; divide both by x^2; 8x or 9 ; 8(111) > 9.
The other - but very simillar - solution is:
On left side we have 111 mutiplications of 222^3 on the right we have 111 mutiplications of 333^2.
Left side is 111^2*111*8 on right side we have 111^2*9.
Since 111 mutiplications of 888 > 111 mutiplications of 9 then left side is bigger:)
"Ou" é o mesmo que "menor maior ou igual", ou seja, um sinal matemático. Tudo q eu fizer de um lado, tomando esse raciocínio, posso fazer do outro.
Oq eu fiz.
222³³³ or 333²²² posso escrever como 222³×¹¹¹ or 333²×¹¹¹. Tirando raiz 111 dos dois lados, cancela, então 111 some dos expoentes😮.
Depois disso, posso deixar a minha ignorância, e tentar tambem escrever 222 e 333 como, (2×111)³ or (3×111)². Meu celular vai descarregar, e n dá pra falar tudo, mas desenvolve isso aí q vai dar certo
Equal
I solved it in 2 and a half minutes.
The biggest power will always produce the biggest number in a sample like this.
222^111 vs (333/222)^222 = (3/2)^222 = (9/4)^111 : very simple
There is an easy way to it.
222^333 = 111^333×(2^3)^111
333^222= 111^222x(3^2)^111
3^2 ~2^3 and 111^333 is waaaay bigger than 111^222, so the option 1 is bigger than option 2
If i'm not stupid, this can be write as 222*222^332 vs 333^222, right? So easy to understand what's bigger.
Have not yet watched the video, but: say X = 222.
Then 1.5x=333
You are now comparing x^333 vs (1.5x)^222. Distributing the exponent 222, you get...
X^333 vs (x^222)(1.5^222).
Divide both sides by x^222...
X^111 vs (1.5) ^222.
The right hand term, however, can be expressed as 1.5^(2*111), which can in turn be expressed as (1.5^2)^111.
X^111 vs (1.5^2)^111.
Raise each side to the exponent (1/111) to get rid of the ^111 part...
X vs (1.5)^2.
Since X is 222, we have 222 vs 2.25. 222>2.25, and thus we arrive at:
222^333 > 333^222.
Just a=x *b
Where a and b are those two numbers then use log on both sides. You get it in 3 steps
A proof by induction would solve all of these in one fell swoop. The general result is far more powerful (and useful).
Divide both sides by 222^222, now you have 222^111 on one side and 1.5^222=2.25^111 on the other. Now it should be clear that the first one is bigger
Just split 222 into 200+22. Then 200^333=2^333×100^333.
Now 333^222=3^222+100^222+33^222.
So directly conude that 222^333>333^222
maths just say if the comparison of two numbers, and their sum of cubic or square root is greater then the outcome number is greater, so start from smaller square, and smaller cube, so , if 222 cube is smaller than 333 squared, the the greater number will be always greater
we obtain 10 941 048 is greater than 110 889
Take log of both side. 333log222 and 222log333. The value of log222 and log333 are between 2 and 3. But 333 is 1.5 times of 222. So 222^333 is bigger.
My head made both exponents to 666 like this:
222^333 or 333^222
sqrt2(222)^666 or sqrt3(333)^666
Then I cancel both exponents and have sqrt2(222) or sqrt3(333). I know the first is just smaller than 15, so take 14. And then check the right side where 14*14*14 is much larger than 333, so the sqrt3(333) is therefore smaller than sqrt2(222). Thus 222^333>333^222
First choice (left)
Take a sip everytime he says "wan wan wan".
333^222. Reasoning: 2^3 = 8, 3^2 = 9. Wrong answer though.. the power of compounding uguuu
So my guess is that for all numbers larger than 1: x^y > y^x if y is bigger than x
e.g. 5^6 > 6^5
And for all numbers smaller than 1: x^y < y^x if y is bigger than x
e.g. 0.5^0.6 < 0.6^0.5
and the closer the numbers are to 1, the smaller the difference
Not true.
2³ is smaller than 3²
8
So much math needed over a simple question wich is just the same as:
Wat is bigger?
2x5 or 5x2...
The only difference is that you use bigger numbers in this problem but the answer is the same.
So you might wanna redo your math here.
After all:
222*333 = We dont know
(111*2 = 222)*333 = Same as line 1
111*333 = (Donno exact number)e+370 = Not even close to being 222? + 2*333 < Also 2x more then on line 2?
So yeah even without calculator i can tell your math is of.
Because on line 2: 2*333 = (with calculator = 3,533694129556769e+72) < i would not know
However on line 3: (2*3*111 = (2x2x2 = 8))x111 = 888? < would know this
Conclusion the lines you write out make no sense in the language of Mathematical algebra.
Even in language of hex code that would be 888136 888136 888136 over code 332193 332193 332193.
Wich roughly translates to 888 and 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000?
Wich makes even less sense to even think they would be near the same.
Conclusion do not add more lines of math to make it more easy,
As it will only make it more easy to make mistakes.
I hope you take this as constructive feedback rather then just a comment that tells you,
That you are wrong.
Because it is not my intention to declare anyone of mistakes.
I only want to point out more learning opportunities for everyone to come to the right conclusion.
(Even if that means i am wrong.)
I hope this helps bring clarity and have a nice day.
222×√222 > 333
both have the same power of 222 so 222^333 is larger... Way larger.
*@Learncommunolizer* -- Here it is using logarithms without guesswork:
222^333 vs. 333^222
333*log(222) vs. 222*log(333)
Divide each side by 111 and expand inside the parentheses:
3*log(111*2) vs. 2*log(111*3)
3*[log(111) + log(2)] vs. 2*[log(111) + log(3)]
3*log(111) + 3*log(2) vs. 2*log(111) + 2*log(3)
log(111) + 3*log(2) vs. 2*log(3)
log(111) + log(8) > log(9)
Therefore, 222^333 > 333^222.
Thanks 🙏❤️🙏
first ones exponent is much larger, and, well, increases exponentially more
This is what I did:
1. I changed 222^333 to (222^1.5)^222 and 333^222 to (222*1.5)^222
2. 222^1.5=222*sqrt 222
3. Since 1.5^2=2.25, and 2.25 333^222
without calculation, you did calculation 1.5^2=2.25
@@gromosawsmiay3000 It's not so hard that you need a calculator for it. 1.5^2=1.5*1.5, and multiplying a number by 1.5 is the same as adding half of the number to itself. Half of 1.5 is 0.75, so 1.5+0.75=2.25.
@@aj_style1745 I do not need calculator, sometimes only pencil and paper, but this was example how to solve this exercise without calculation. I remember my time on technical university.... I solved integrals in my memory when I travel by bus :-)
2^3=8