X=(1/y)^2 so original becomes (1/y^2)^y=1/16 So y^2y=16=2^4 So let f(y)=y^(2y) then f'(y)= y^2y*ln(y)+y^2y which is 0 when lnx=-1 . Since y^2y is always positive then the derivative changes sign only at y=e^-1. Since lim(y->0) y^2y=1 we only have the obvious solution on the right side when y=2 Going back x=1/y^2=1/4
I can't with this guy. He has a main channel for videos, creates a secondary channel specifically for shorts and then uploads long videos on it. Why'd you even called it "SyberMath Shorts"?
Thanks Sir 👍
_What's the word?... Vertical tangents._
I.e., vertical asymptotes.
All very thoroughly worked out, thank you.
X=(1/y)^2 so original becomes
(1/y^2)^y=1/16
So y^2y=16=2^4
So let f(y)=y^(2y) then f'(y)= y^2y*ln(y)+y^2y which is 0 when lnx=-1 . Since y^2y is always positive then the derivative changes sign only at y=e^-1. Since lim(y->0) y^2y=1 we only have the obvious solution on the right side when y=2
Going back x=1/y^2=1/4
1/16=(¼)²
Change the exponent: 2=1/½
=1/sqrt(¼)
x=¼
x = 1/4
1/x^(x)^1=2=1/16 1/4^(4)^1/2
1/4^2=1/16 x=4, or
1/2^2^2=1/16 x=2^2
(1/4)^2 i. e x= 1/4
X=1/4
I can't with this guy. He has a main channel for videos, creates a secondary channel specifically for shorts and then uploads long videos on it. Why'd you even called it "SyberMath Shorts"?
4 and -4?
Wrong and Wrong?