@@barberickarc3460 Actually I did. And it said has no solution. The case is that I watched a video in another channel saying there would be many complex solutions like here, all steps seemed correct. So I got confused. But thanks for your reply anyway
I think the confusion is coming from the definition of exponentiation. In complex analysis, z^w is defined as exp(w*log(z)), where log(z) is multi-valued. If you use that definition for 1^x, you get exp(x*log(1)), but then you have to decide what log(1) is equal to. You could say it is multivalued: log(1) = 2*pi*i*n , and then you can write a solution for 1^x = 2. Technically, infinitely many solutions. But unless you are solving some esoteric problem, that is not the usual assumption. Outside of complex analysis class, most people assume that z^w is NOT multi-valued, and the principal value is used.
It bothered me that you only chose one branch of it but in the end, the left hand side would have an extra multiple of 2π which can just be absorbed into the multiple on the right hand side.
You must write all possible angles by adding the 2pik term to pi/2. Because after simplifying you will see it doesn’t just vary by a integer multiple of 4 but rather it boils down to adding an even integer divided by an odd integer. That means multiples of 4 added to it are not the only solutions but rather ratios of even over odd integers.
@@SyberMath he’s also wrong Syber it does work. That commenter made a mistake by saying i= pi/2 +2kpi. That is not true. That is true for e^i(pi/2 +2kpi).
A lot of solutions are missing i = pi/2 + 2k pi (where k is an integer) so the solutions are : x = 4n/(4k+1) - i 2ln2/((4k +1) pi) (where k and n are integers) Of course, in the video, the only solutions you get are those with k = 0
No, he did not miss any solutions. Your "solutions" do not solve the equation except for k = 0. If you try n=0, k=1, you will see that they do not satisfy the original equation.
You need to learn how to define the complex log function so it is single- valued. Your definition says ln(1) = 2 (pi ) i for any integer n and not just 0.
With due respect, sir, your calculations with complex logarithms are WRONG. You need to undestrand the Riemann surface of the logarithm. (Google it!) An exanple: exp(i pi) = - 1 is a well known quation . But we also have: exp(- i pi) = - 1. Naively taking logs here, without understanding the Riemann structure, would give the result i pi = - i pi, which clearly is not true. The mistake is to ignore that the usual definition of the log function is such that it has an imagiary part equal to or less than pi and bigger but not equal to - pi. Othewise I enjoy your stuff. Ebbe Nyman
Yesterday you posted i^x=1 as a short, so this seemed pretty easy to solve.
Nice!
I wish there were a way to get that pi out of the denominator
Bravo
Très sympa
Merci
Nice, clear explanation, as usual.
Glad you think so!
@@SyberMath why not just take natural log on both sides at beginning and then solve for x..you can do it so y not just curious?
@@SyberMath and why not just keep 2 as 2..i see novalue inmultiplying by e^i2npi
My question is, we take the general form of 1 with all possible angles, but we don't for i. Why is that.
I got x = [2ln(2)] / (iπ) quite quickly, but forgot the general form, very nice problem
Thanks
Amazing question with the best solution...👍👍👍.
Many many thanks 😍
I love math
For a second method the superpower of substitution can by used by replacing x by a+bi
Thank you for the videos, these are really fun to solve
Glad you like them!
@@SyberMath :)
x=log(i)2=ln2/lni=-iln4/pi
is there any solution for 1 ^ x = 2?
there is not. 1 raised to any number x where x ∈ ℂ is 1. For questions like this you can check wolfram alpha for a quick answer
@@barberickarc3460 Actually I did. And it said has no solution. The case is that I watched a video in another channel saying there would be many complex solutions like here, all steps seemed correct. So I got confused. But thanks for your reply anyway
I think the confusion is coming from the definition of exponentiation. In complex analysis,
z^w is defined as exp(w*log(z)), where log(z) is multi-valued. If you use that definition for
1^x, you get exp(x*log(1)), but then you have to decide what log(1) is equal to. You could say it is multivalued: log(1) = 2*pi*i*n , and then you can write a solution for 1^x = 2. Technically, infinitely many solutions.
But unless you are solving some esoteric problem, that is not the usual assumption. Outside of complex analysis class, most people assume that z^w is NOT multi-valued, and the principal value is used.
@@guilherme.siqueira no idea what went wrong in the video but if you actually plug in the value of x = [ln(2)] / 2iπ] you get 1^x = 1
It bothered me that you only chose one branch of it but in the end, the left hand side would have an extra multiple of 2π which can just be absorbed into the multiple on the right hand side.
You must write all possible angles by adding the 2pik term to pi/2. Because after simplifying you will see it doesn’t just vary by a integer multiple of 4 but rather it boils down to adding an even integer divided by an odd integer. That means multiples of 4 added to it are not the only solutions but rather ratios of even over odd integers.
See @XJWill1's comment...
@@SyberMath he’s also wrong Syber it does work. That commenter made a mistake by saying i= pi/2 +2kpi. That is not true. That is true for e^i(pi/2 +2kpi).
لايمكن تطبيق قانون موافر
Good solution. I did this and got no solutions:
Square both sides to get
(-1)^x = 4
And again
1^x = 16
x ln(1) = ln(16)
0 = ln(16)
Anyone???
when n is not integer (a^m)^n≠a^(mn), please see my other comments for it
Hallo everytwo
For me i will make it x=ln( 2)/ln(i) then ln (i)=iπ/2 so the answer is
X= -i ln(4)/π😅😅😅
A lot of solutions are missing
i = pi/2 + 2k pi (where k is an integer) so the solutions are : x = 4n/(4k+1) - i 2ln2/((4k +1) pi) (where k and n are integers)
Of course, in the video, the only solutions you get are those with k = 0
That would be (e ^ i) raised to that power, not as you wrote it.
No, he did not miss any solutions. Your "solutions" do not solve the equation except for k = 0. If you try n=0, k=1, you will see that they do not satisfy the original equation.
@@XJWill1 why is that, so what's wrong with maitrehenryalain7765 idea?
Thank you! 😊
@@femtogary3723 What do you mean? The answer he gave does not satisfy the given equation unless k=0.
You need to learn how to define the complex log function so it is single- valued. Your definition says ln(1) = 2 (pi ) i for any integer n and not just 0.
not defined. i^2 = -1
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With due respect, sir, your calculations with complex logarithms are WRONG. You need to undestrand the Riemann surface of the logarithm. (Google it!) An exanple: exp(i pi) = - 1 is a well known quation . But we also have: exp(- i pi) = - 1. Naively taking logs here, without understanding the Riemann structure, would give the result i pi = - i pi, which clearly is not true. The mistake is to ignore that the usual definition of the log function is such that it has an imagiary part equal to or less than pi and bigger but not equal to - pi. Othewise I enjoy your stuff. Ebbe Nyman