A Nice Exponential Equation | iˣ = e

[@scottleung9587]

Got Syber's answer - WA did its arithmetic a little differently, but if you think about it you'll pretty much get the same values for any integer n.

[@mathsandsciencechannel]

To solve the equation "imaginary to the power x equals e," we can rewrite it using the exponential form of complex numbers. The imaginary unit, denoted as "i," can be expressed as e^(iπ/2) in exponential form. Therefore, the equation becomes: (e^(iπ/2))^x = e Now, we can use the property of exponents, which states that when we raise a power to another power, we multiply the exponents: e^(ixπ/2) = e Since the bases (e) are equal, we can equate the exponents: ixπ/2 = 1 To isolate x, we'll divide both sides of the equation by iπ/2: x = 1/(iπ/2) To simplify the expression further, we can multiply the numerator and denominator by 2/iπ: x = (2/iπ)/(iπ/2) = 4/(i²π) The value of i² is -1, so we can substitute it into the equation: x = 4/(-1 * π) = -4/π Therefore, the solution to the equation "imaginary to the power x equals e" is x = -4/π.

[@aintaintaword666]

Hello ChatGPT

[@mcwulf25]

Pretty much what I did

[@davibz2243]

"Since the bases (e) are equal, we can equate the exponents:" - No, you certainly can't do that. Are you a bot or something?

[@mathsandsciencechannel]

@@davibz2243Why?

[@mathsandsciencechannel]

@@davibz2243 The euler e is a constant and has a value so it just like having any number as base on both sides

[@Paul-222]

I used the alternate form from WA as my first step, then set the exponents equal to each other. It seemed more direct. I only got the principle solution that way, but it’s all I was after.

[@someperson188]

The difficulty is that non-integral exponents aren't well-defined in the complex numbers. For example, what is i^(2/3)? Since i = e^(πi/2) = e^(πi/2 + 2πi), we can define i^(2/3) in at least the following two ways: e^[(πi/2)(2/3)] = e^(πi/3) = cos(π/3) + i sin(π/3) = 1/2 + (sqrt(3)/2)i. e^[(πi/2 + 2πi)(2/3)] = e^(5π/3) = cos(5π/3) + i sin(5π/3) = 1/2 - (sqrt(3)/2)i. It's easily seen that there are 3 different possible answers by this "definition", which can be obtained by multiplying e^[(πi/2)(2/3)] by the three complex cube roots of 1. If x is irrational, the same idea gives infinitely many values of i^x.

[@mcwulf25]

I didn't use ln. Took the xth root of both sides and turned i into the Euler form as you did. Then compared exponents (ok, effectively taking ln of both sides) to get the answer

[@RexxSchneider]

We can write i^x as e^(2nπi + πi/2)^x, where n is an arbitrary integer. But, we can also write e = e^(1 + 2mπi), where m is another, INDEPENDENT, arbitrary integer. Equating exponents, or taking natural logs, we get (2nπi + πi/2)x = (1 + 2mπi). So x = (2mπi +1) / (2n + 1/2)πi = (2m + 1/πi) / (2n + 1/2) = (4m - 2i/π) / (4n + 1) or (4mπ -2i) / (4n+1)π. When m = 0, we have your solution, but here are an infinite number more such as (4π - 2i) / (4n +1)π when m = 1, etc.

[@gauchisteanonyme3684]

Wolfram Alpha gives the right answer. Your answer is incorrect. Just check your solution with n=1, i^x = e^(1/5) and not e.

[@MikeMagTech]

I recently subscribed to your channel and am very glad I did!

[@SyberMath]

Awesome! Thank you!

[@derhausfreund5691]

log i=log | i | + i *pi/2 usw.

[@GirishManjunathMusic]

i↑x = e Taking both sides to the power of iπ: (i↑x)↑iπ = e↑iπ i↑iπx = e↑iπ As e↑iπ = 1, i↑iπx = 1 But as 1 = i↑4n, (where n is any positive integer) i↑iπx = i↑4n Comparing exponents: iπx = 4n x = 4n/iπ; where n is any positive integer.

[@omnipresentcatgod245]

e^iπ is -1 not 1

[@ThomasLoganRitchie]

i^x=e --> exp(x log(i) ) = e --> exp (x iπ/2) = e --> x iπ/2 = 1 + 2hπi --> x = -(2/π)i + 4h, for any integer h.

[@usdescartes]

Ah. I see the problem. Before taking the log of both sides, you MUST change e into e•e^(i2πn). In fact, it doesn't matter what form you use for i, since x will always be the exponent of some base that equals i. It's i no matter what form you want to make it, so you don't need to change it into a complex (LOL) Complex polar form.

[@fabriziosantin6063]

Shouldn't one have to consider ln(e) = 1 + i(2πk) with k integer, instead of just 1?

[@RexxSchneider]

Indeed one should. x = (4kπ - 2i) / (4n + 1)π.

[@usdescartes]

@@RexxSchneider The main issue, in case I haven't been clear, is that i = i = i, and it doesn't matter what form it's in. And we are trying to find how to get i^x to equal e. Now, there are lots of forms e could be in, too... but these are the ONLY ones we care about, because different values of x will give us different forms of a Complex number, and when x = (4k + 2i/π), we cover all possible Complex forms for e. Why does that matter? First, no matter what Complex form i has, our exponent x must make it so that i^x = e. Your form with the (4n + 1) in does NOT give us e no matter what form i is in. Second, if we have covered every possible value of x that gives the result of e, then that's our answer. No other manipulation will make the answer "better." It's rather easy to show that x = (4k + 2i/π) gives us e, no matter what k is, so we know that this is the answer. Not only that, but the only way to make 4k + 2i/π = (4kπ - 2i)/((4n + 1)π) is to have n = 0. Not only THAT, but it's also easy to show that your x will give us i^x = e^(n/5)... which never equals e unless n = 0. ¯\_(ツ)_/¯

[@RexxSchneider]

@@usdescartes I follow your logic and I agree. You make one mistake, in that if x = (4kπ - 2i) / (4n + 1)π, then i^x = e^(1/(4n+1)), although the conclusion that n has to be zero remains.

[@RexxSchneider]

@@usdescartes On further reflection, you almost caught me out. I accepted your claim that _" i = i = i, and it doesn't matter what form it's in"._ That turns out to be untrue, and you can see that quite clearly when you use the form i = e^(5πi/2) as an example. Using your value for x = (4k - 2i/π ), we get: i^x = e^(5π i/2)^x = e^(5π i(2k - i/π )) i^x = e^(10kπ i - 5i^2) = e^5 That's not e, and your solution for x does not work for all forms of i. If you use my value for x = (4k - 2i/π )/(4n+1), we find that: i^x = e^((5π i/2)(4k - 2i/π) / (4n + 1)) i^x = e^((10kπ i + 5) / (4n+1)) i^x = e^(5 / (4n+1)) and you choose n=1 to get e as the answer when using that form of i. Similarly, we can use any representation of i of the form e^i(2n+1/2) and then my solution x = (4k - 2i/π )/(4n+1) -- where the n is the same in each -- and we will arrive at the answer e. Hope that helps you understand.

[@charlesgodswill6161]

At what point do we impose the principle of n=0 ?

[@kassuskassus6263]

Yesssssss !!!!!

[@albertoalves1158]

May I suggest to take both sides to the power of 2. i^2x = e^2 but: i^2 = -1 = e^i.pi Then: e^i.pi.x = e^2 Then: i.pi.x =2 x = 2 / i.pi = -2 / pi. !!! Any mistake ???

[@RexxSchneider]

Yes. When you square both sides of an equation, you may introduce spurious invalid solutions. For example: x = -1 so x^2 = 1, so x = 1. So care is needed. However, in your case, you wrote: x = 2 / iπ which is actually equal to -2i / π. In other words, you left out i from your final result. You found one of the solutions, which corresponds to n=0 in the solution in the video. I believe the full solution is x = (4mπ - 2i) / (4n + 1)π, where m and n are arbitrary, independent integers.

[@usdescartes]

@@RexxSchneider It's easy to show you solution doesn't work. Just plug it in and check. The reason it doesn't work is that the (4n+1) part you have for x ONLY works for very specific forms of i. That is, if you want to use n = 1, you get x=(4m - 2/(5π) i). So iˣ = i⁴ᵐ∙i^(2i/(5π)) = 1∙i^(2i/(5π)) = e^(1/5) ≠ e. Oops. The deal here is, you MUST be careful if you change terms on BOTH sides of a Complex equation you are about to take the logarithm of. Just like you can get false answers when you square both sides, you can get false answers, as you have demonstrated, when you change numbers on both sides to polar Complex form and take a logarithm. Here's a nice, written-out demonstration for you: twitter.com/nandor117/status/1668679775852871680 If you haven't seen the real answer yet, it's here: twitter.com/nandor117/status/1668679524924399627

[@protoman1365]

My guess: x = 2/(pi*i), or x = -2i/pi Reason: e^i*pi/2 = i. Simply just raise both sides to the power of 2/(i*pi) and you’re done. Edit: after watching the video, it seems like my logic would have worked, but the full solution would have come if I instead started with e^i*(4n+1)*(pi/2). I just took the first easy solution.

[@giuseppemalaguti435]

x=1/lni

[@Erudecorp]

logᵢ(e)

[@Alexey_Alex1]

x=1/ln(i)

[@mathsandsciencechannel]

Am about to premiere guys. Thanks for the solution @sybermath