Help me solve for X nearly impossible but still possible I don't know how

[GreenPrint]

Help me solve for X nearly impossible! but still possible I don't know how

Homework Statement

Any help you can offer would be great for solving this problem. I got very far in the solution and just kept on running into dead ends. Multiple forms of the equaiton below that I was clueless on how to solve.



If you help me solve this, that would be crazy. THANK YOU if you do as I don't see how but maybe it is possible my teacher is so mean making me solve this problem. Seems like she's trying to be mean all the instruction were 5 words:

"Solve"
and the problem in the picture above.
O.M.G. I don't know how to do this problem.

Homework Equations

The Attempt at a Solution

 

[iamalexalright]

well, if I read the problem correctly isnt:

i^{2x/\pi} - i^{2x/\pi} = 0 ??

 

[GreenPrint]

crap yah =( I put it wrong forgot the negative
http://img683.imageshack.us/img683/8587/capturefw.jpg

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that should do it I'm really sorry about that

 

[GreenPrint]

It is possilbe to solve though correct I don't see why not you just need to know how which i don't know how to do I got to several dead ends

 

[vela]

It would really help if you used proper punctuation.

This problem can't be solved exactly. You can simplify it a bit if you use the fact that i=e^{i\pi/2} in the numerator of the lefthand side and do some algebra. It's clear from the resulting equation, however, that you can't find an exact answer.

 

[GreenPrint]

and why is that? Why can't you fidn an exact answer

 

[GreenPrint]

I think if you just knew how to, which I don't know how to, you could solve it, as I really don't see why it's no solvable can you maybe explain more why it isn't solvable

 

[vela]

You can't solve it exactly for the same reason you can't solve an equation like x^2e^x = x+10 exactly. You can get an approximate solution using numerical methods, but you can't write down a solution in terms of elementary functions.

 

[Pengwuino]

Here is a hint as to why it is unsolvable.

An Euler identity is e^{ix} = cos(x) + isin(x). Substitute that in for what you have and you'll quickly see why it isn't solvable exactly.

 

[GreenPrint]

Yes but trigonemtric functions can take on an infinite amount of values so to say that it has no solution bothers me becasue the solution can consist of all possible numbers possilbe, all of them, to say that there is no solution seems to be as bad as teaching people that cos(x) = -2 has no solution because it has an infinite amount of solutions

http://img6.imageshack.us/img6/7261/capturefdr.jpg

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so i do believe their is one I just don't kno whow to solve that equation for X

 

[GreenPrint]

Actually graphing it does reveal a solution I was right

 

[GreenPrint]

http://www.wolframalpha.com/input/?i=((i+2x/%CF%80-i+(-2x)/%CF%80))/2i%3Dx^3-27,+solve+for+x

see that page there is a solution so now that you believe me can you please help me solve thanks

 

[cristo]

http://www.wolframalpha.com/input/?i=((i+2x/%CF%80-i+(-2x)/%CF%80))/2i%3Dx^3-27,+solve+for+x

see that page there is a solution so now that you believe me can you please help me solve thanks

The equation you typed into wolfram|alpha here, and obtained a result for, is not the same as the equation you give in your opening post (you missed out the exponents, and wrote them as multiplications). Putting the correct equation into wolfram|alpha returns an approximate answer (since, as you have been told, this cannot be solved exactly):

http://www.wolframalpha.com/input/?i=%28i^%282x%2FPi%29-i^%28-2x%2FPi%29%29%2F2i%3Dx^3-27%2C+solve+for+x

 

[GreenPrint]

hmmm Thank You

 

[vela]

I didn't say there was no solution. I said you can't solve it exactly.

 

[statdad]

I think vela is on to something here: I think the original poster is saying "solve it" with the intent of writing a solution explicitly as x = some expression, while others have been using "solve it" as a request to find any number(s) that work, realizing that there is no closed form for those values - i.e., found approximately.