e^(i*pi)=-1 but e^(3i*pi) is also -1. And that gives another solution: ln5 / (3i*pi +ln5) The same if we replace 3i by 5i, 7i ,9i, ... So the solution in the video is not complete.
I think you can make it a lot simpler by doing: (-5)^x=5 x=log_{-5}(5) x=ln(5)/ln(-5) x=ln(5)/(ln(-1)+ln(5)) Then you can turn e^(i*pi)=-1 into ln(-1)=i*pi and substitute that in to get: x=ln(5)/(i*pi+ln(5))
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e^(i*pi)=-1 but e^(3i*pi) is also -1. And that gives another solution: ln5 / (3i*pi +ln5)
The same if we replace 3i by 5i, 7i ,9i, ...
So the solution in the video is not complete.
I think you can make it a lot simpler by doing:
(-5)^x=5
x=log_{-5}(5)
x=ln(5)/ln(-5)
x=ln(5)/(ln(-1)+ln(5))
Then you can turn e^(i*pi)=-1 into ln(-1)=i*pi and substitute that in to get:
x=ln(5)/(i*pi+ln(5))
you beat me to that.
Just FYI the name Euler is actually pronounced like "oy - lur." This video is super cool, thanks!
In what programm do you drawing?
Why can it not simply be 2/2, 4/4, or any other even number divided by itself? Sqrt (x squared) = |x|
In pure mathematics the sqrt of a number has two possible answers
ex: √9 = ±3
Meaning the square root of 9 is both 3 and -3
@theevilmoppet Because the division in the exponent would happen first, so (-5)^(2/2) = -5.
The solutions are as follows :
X = ( ln(5)^^2 + 2k.(pi^^2) +
i.(2k-1).pi.ln(5) )
/
( ln(5)^^2 + pi^^2 )
with k € Z.
Excellent! 👏