Math Olympiad Question | Compare 100^99 and 99^100 which is larger?

[benbooth]

If one has a problem in the form A^B=B^A, and both A and B are greater than e, then the term with the largest exponent is always the largest.

My solution before watching the video

100^99 ≠ 99^100
100^99 / 99^100 ≠ 1
100^99 / 99 * 99^99 ≠ 1
(100 / 99)^99 * 1/99 ≠ 1
(1+99 / 99)^99 * 1/99 ≠ 1
(1 + 1/99)^99 * 1/99 ≠ 1

Now...
(1 + 1/99)^99
is in the form...
(1 + 1/n)^n = f(n).
The limit of f(n) as n approaches infinity is e. Therefore we can say that...
(1 + 1/99)^99 * 1/99 < e/99

Therefore...
(1 + 1/99)^99 * 1/99 < e/99 < 1
As the fraction 100^99 / 99^100 has a value less than 1 we can conclude that 100^99 < 99^100.

Same method!

[Dr_piFrog]

Just compare the log of both numbers and compare.

[soumiscreativecorner]

Good night dear friend 💖💖👌👌important vedio

[umergulzar]

Good I need help with this question. (√x)^2^(1/2)=(1/2)^2^(√x). I know the solution is 1/4 by hit and trial method... But i want a step by step solution...

[MrRandycrum]

a^b > b^a for all a and b > 2 and 3. I have not proved this yet. But I am pretty sure.

[ajaypratap]

Have yiu ever made the graph of ( 1 + 1/x ) ^x ... What is the value it converges to for large numbers ? Does it even converges ?