we can log both side first: (logx)*log(lnx)=log(x^2) (logx)*log(lnx)-2logx=0 (logx)[log(lnx)-2]=0 We have 2 solutions: logx=0x=1 log(lnx)=2lnx=10^2=100 x=e^100
It would be interesting if you could "map" this problem to aplusbi for x in the complex domain. Some might say it’s an un"real" thought. In any case keep cooking them up.
Not sure i am readig this right but doing base transformation logx =lnx/ln10 and substituting x=e^y we obtain: Y^(y/ln10)=e^y^2 Raising to power ln10 Y^y=10+e^y^2 Gonna watch the video and see if i understand the equation properly
Another nice problem that i solved in my head,keep up the great work syber!😊💯
Thanks, will do!
@@SyberMath video request
Integral (1-sinx)^1/2 from 0 to pi
Hint: u -sub x=2theta
Can u come up with other methods?
we can log both side first:
(logx)*log(lnx)=log(x^2)
(logx)*log(lnx)-2logx=0
(logx)[log(lnx)-2]=0
We have 2 solutions:
logx=0x=1
log(lnx)=2lnx=10^2=100
x=e^100
Easiest way is to present right side as (10^ logx)^2 which is equal to 100^
logx. Then both sides are to the same power, hence lnx = 100
It would be interesting if you could "map" this problem to aplusbi for x in the complex domain. Some might say it’s an un"real" thought.
In any case keep cooking them up.
Not sure i am readig this right but doing base transformation logx =lnx/ln10 and substituting x=e^y we obtain:
Y^(y/ln10)=e^y^2
Raising to power ln10
Y^y=10+e^y^2
Gonna watch the video and see if i understand the equation properly
Nice!
Thanks!
This is homemade. You sure know how to cook up a beaut!
Thank you, Robert! 😍
Can I solve it? Yes, I can. AND, I did not need to invoke Lambert's W function, so there!
Not convinced about x = 1.
Defo 0 power to 0 is not 1
Well google says yes it is sometimes. So...the answer is x=1
ruclips.net/video/kv5VY0REbMg/видео.html
ruclips.net/video/kv5VY0REbMg/видео.html
(ln x)^(log x) = x^2
log((ln x)^(log x)) = log(x^2)
(log x)*log(ln x) = 2 log x
(log x)*(log(ln x) - 2) = 0
log(ln x) - 2 = 0
log(ln x) = 2
ln x = 10^2 = 100
x = e^100
log x = 0
x = 1 (if one accepts 0^0 = 1)
It is.
x value is huge.... x = e^100... l haven't calculated the value, but it is...
x = 1 is a obvious solution
You get an indeterminate form with x = 1, not a solution.
@@doctorb9264 you are right.
e°100
Can "i" solve this? Yes "i" can but only in my "imaginary", un"real" world!!!
u and i can solve this together! 😍😁
@@SyberMath: u-bet!
u +(and) i is just as good as a + bi!
ln(x)^log(x)=x²
(ln(x)^log(x))^(log_x(10))=(x²)^(log_x(10))
ln(x)^log(10)=100
ln(x)=100
x=e¹⁰⁰
I got the e^100 but missed the obvious solution.
Don’t worry if you didn’t. 0^0 is an indeterminate form. blackpenredpen just posted a (second) video showing that 0^0 be 0 today.
He's talking about limits not the value of 0⁰
1st
x ∈ { 1, e¹⁰⁰ }