Question

i^x=x

Answer 1

Help!

Answer 2

What are you trying to do?

Answer 3

Solve for x

Answer 4

is there an other way than hitting it with a log

Answer 5

@RadEn

Answer 6

@wio

Answer 7

So you can't use natural log?

Answer 8

No

Answer 9

A log doesn't seem helpful...

x log(i) = log x isn't really any more solved than before.

Answer 10

http://www.wolframalpha.com/input/?i=i%5Ex%3Dx+solve+for+x

i have no idea other than that.

Answer 11

So how do I solve that (w/o log)?

Answer 12

This is \(i\) as in \(\sqrt{-1}\) right?

Answer 13

yes

Answer 14

With a natural log you can make it slightly less atrocious, but without using it I have no idea. Even with using it I have not many ideas.

Answer 15

x can;t be zero either, as I see, right?

Answer 16

\(i^0 \neq 0\) so no

Answer 17

Ok

Answer 18

You could clean it up a little bit using Euler's Identity,\[\Large e^{\frac{\pi}{2}i}\quad=\quad i\]So we can write this as:\[\Large e^{\frac{\pi}{2}ix}\quad=\quad x\]Which relates somehow to that weird product log function.. I don't really know much about it myself though D:

Answer 19

@zepdrix, i don't understand that, sorry, can you explain?

Answer 20

I have no idea how to solve this by hand, so i'm wondering what class it's from...

Answer 21

e is 2.7 , right?...but still..

??????

Answer 22

\[\Large e^{ix}\quad=\quad \cos(x)+i \sin(x)\]\[\Large e^{i\frac{\pi}{2}}\quad=\quad \cos(\pi/2)+i \sin(\pi/2)\]\[\Large e^{i\frac{\pi}{2}}\quad=\quad 0+i (1)\]That's what I meant by Euler D:

Answer 23

@agent0smith, I posted in a wrong section maybe, but, isn't this from algebra 1 just that with i?

Answer 24

This is way beyond algebra 1 stuff

Answer 25

why that?
how do you get that I am completely lost!

Answer 26

What level of math is this for solomon? XD

Answer 27

I don't really know, think trig or alg 1 or alg2, not sure

Answer 28

lol whut? :D Did you just pull this problem out of your butt?

Answer 29

LOL algebra 1 doesn't involve ANYTHING like this. Neither does algebra 2, or early calculus. I'd guess maybe some complex number theory classes might.

Answer 30

No. I was looking at tough SAT problems online....

Answer 31

My first guess was no solution, but I am not sure.

Answer 32

Yeah SAT problems are possible to solve by hand without some weird product log function.

Answer 33

Tough SAT problems are doable by bright high schoolers. Pretty much the only high school doing this one is one of those guys who proves Fermat's last theorem when he's 20

Answer 34

SAT also doesn't require knowledge of imaginary numbers.

Answer 35

OK than it was just tough math problems I was looking both sat and reg/ non-sat and wrote them down on my paper , don't know which is which...

Answer 36

There is a solution. Wolfram Alpha gives it to you. It does not offer a step by step solution, though, so I have no idea where it got it.

Answer 37

I have some others maybe you can solve them for me? If you can't solve this.

Answer 38

I wouldn't even know how to solve this if it was 2^x = x but at least you can graph it... it's even weirder with an i in it's place.

Answer 39

@SACAPUNTAS, I agree, but as I already said, I don't understand that solution, so I asked to explain.

Answer 40

@agent0smith, it is weird, but.... can you solve it?

Answer 41

I'm sure there's someone on th the site who can explain it... maybe.

Answer 42

@wio maybe?

Answer 43

I a;ready mentioned him b/f

Answer 44

So none of you guys can help?

Answer 45

I can't! I was just hanging around in hopes that someone could explain it. :D

Answer 46

I see many PROs and no help yet.....

Answer 47

I'd wonder if anyone on the site can help. You're asking a question which is above most people's level of math knowledge. I've never used the product log function which seems required.

Answer 48

Sorry, but I am maybe smarter than asking easier questions,

Answer 49

Whatever.... HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 50

RECALLING THE PROBLEM AGAIN



i^x=x

Answer 51

Maybe