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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!!!
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
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 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
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.
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.
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
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 :(
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.
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
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
"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
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
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.
*@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.
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
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
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.
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:)
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
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.
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.
how I did it, 222^333 (=) 333^222 ( (=) currently unknown inequality) 222^(1/222) (=) 333^(1/333) then for x>e (Euler's number) x^(1/x) gets smaller so 222^(1/222) > 333^(1/333) therefore 222^333 > 333^222
Uhh... When in a multiplication the exponents are the same but bases are different, you can't just multiply the bases and keep the exponents the same.. Thats not how it works.. Like, using a simple example, if its 2^2 × 3^2 then the answer would be 13.. And when you just multiply the bases and put the power the same, which is what you did for 111^111 × 8^111, it is not correct as it would just be 6^2 which is 36..13 and 36 are not the same.
There are many of these videos going around: compare a^b and b^a. Well, to simplify, look at b log a ??? a log b, or (log a)/a ??? (log b) / b. Or, generally consider the behavior of the function f(x) = (log x)/x. By quotient rule, f'(x) = [x(1/x) - log x] / x^2 = (1-log x)/x^2. So, f'(x) > 0 for x < e, and f'(x) < 0 for x > e. In this case, e < 222 < 333, so log(222)/222 > log(333)/333, i.e., 333 log 222 > 222 log 333, i.e. 222^333 > 333^222.
