These problems are quite easy for me. If the numbers are greater than or equal to e (2.718...) then the one with the smaller base would always be the bigger one.

if e <= a < b then a^b > b^a

1. Take the 111th root on both numbers
222^333 -> 222^3
333^222 -> 333^2
222^3=10,941,048 , 333^2=110,889

0:16 why would ANYONE use a calculator for this,if both the exponentials you are comparing have a base bigger than e then the one with the smaller base and bigger exponent will always be bigger

A very common math puzzle, like e^pi or pi^e where you divide both powers by e*pi and then bring to a function f(x)=x^(1/x) and find derivative and equate it to 0. SO you find out that its maximum is at e and after that it is decreasing. So any smaller base that is > E will always win, thus 222^333 is bigger.

Please continue your integration series. You have only told about two methods of integrating. And you have told a very small about parts integration. Please make videos on other methods❤.

Congratulations great explanation for sure 👍😀

Great explanation 😊

Nice explanation and stuff. But I know what if the power or expotant is bigger, the answer will also be bigger. Only reason I watched is cuz I was wondering if the expotant is big enough

Congrats!!! Great Explanation!!! Thank You Very Much. May GOD still BLESS You. I’m from Brazil 🇧🇷.

Consider the general case x^(x+y),and (x+y)^x,with  x > y>0 We want to show that
x^(x+y) >(x+y)^x. To this end we consider the logarithms:(x+y)*ln x > x*ln(x+y).
ln(x+y) =ln (x*(1 + y/x)=  ln(x) +ln(1+y/x).  We can use the fact  that
ln(1+ y/x ) < y/x . If we use this for the logarithms
we find that x^(x+y) > (x+y)^x if y*ln (x) > y , i.e. x ≥ e .

How to figure out which is larger, you have to compare what number is larger in the equation 222log e (333( vs the other way around

A calculator will be helpful. Take log 222 and log 333. Multiply both logs by 333 and 222, and see which log is larger.

3 squared is bigger than 2 cubed. Easy as that. Everything else is bullshit.

i tried like this “23 vs 32”, got correct in a nanosecond

222³³³ = 222×222×222...×222×222 (333 times)
Wehre 333²²²= 333×333×333...×333×333 (222 times)
The first one is much larger because it's repeat a hundred times than the second one.

Bro better exponent means greater value that's why 222^333 is greater

So when it involves numbers way in excess of the number of atoms in the universe, the real answer is it doesn’t matter

Bro if 222=x and 333=y

Then the no. with y as exponent will surely be larger than the no. with x as exponent if the bases are interchanged, except for one condition if the bases and powers are 2 and 3 or numbers smaller than them.

Express terns in of x=111 the
2x^3x vs 3x^2x
(8x³)^x vs (9x²)^x
Set inequality,multiply both sides by 1/x and then 1/x²,see that
8x>9
X>9/8(greater than 1)
Since 111=x
Then it holds such that 222³³³>333²²²