An Exponential Equation with Logs | Surprising Results

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[@Mohammad1Ahmad]

Thank you very much, a professional teacher who deserves all respect and appreciation

[@XJWill1]

I would like to know what you typed into Wolfram Alpha to get that set of solutions, because when I type :
complex solutions z^(Log(z)/Log(10)) = (sqrt(z))^Log(z)
or anything like that, it gives me a single solution, z = 1. Which, incidentally, agrees with my own computation, assuming everything is taken to be the principal branch (principal value) in the given equation. There are no complex solutions in the principal branch of the equation shown in the video. And since WA uses the principal branch, it should not have returned the solutions you showed.
By the way, I have commented repeatedly that your misapplication of exponentiation identities that are valid for positive-real-numbers but you apply them to complex-valued numbers, will get you into trouble eventually. Your answer in this video is incorrect. And (z^w)^p does NOT equal z^(w*p) in general for complex numbers (or even for some negative real numbers).

[@XJWill1]

Here is a solution over the complex numbers, assuming the principal branch of every function in the equation given in the video. Spoiler: no non-real valued solutions exist
Note that Log() is the principal branch of the complex-valued natural Logarithm function
PV: z^(Log(z)/Log(10)) = (sqrt(z))^Log(z)
exp(Log(z)*Log(z)/Log(10)) = exp(Log(sqrt(z))*Log(z))
exp(Log(z)*Log(z)/Log(10)) = exp(1/2 * Log(z)*Log(z))
Note that Log(z^x) = x*Log(z) for x real-valued and abs(x) < 1
exp(Log(z)^2 * (1/2 - 1/Log(10))) = 1
log( exp(Log(z)^2 * (1/2 - 1/Log(10))) ) = log(1)
Log(z)^2 * (1/2 - 1/Log(10)) = i*k*2*pi any integer k
Log(z)^2 = i*k*2*pi / (1/2 - 1/Log(10))
Log(z) = (-1)^m * ( i*k*2*pi / (1/2 - 1/Log(10)) )^(1/2)
Log(z) = (-1)^m * (i*k)^(1/2) * sqrt(2*pi / (1/2 - 1/Log(10)) )
Note that (w*z)^p = w^p * z^p if w or z are positive real-valued
Log(z) = (-1)^m * sqrt(abs(k)/2) * (1 + i * sgn(k)) * sqrt(2*pi / (1/2 - 1/Log(10)) )
= (-1)^m * sqrt(abs(k)) * (1 + i * sgn(k)) * sqrt( pi / (1/2 - 1/Log(10)) )
= (-1)^m * sqrt(abs(k)) * (1 + i * sgn(k)) * 6.9147...
Note that -pi < Im( Log(z) )

[@SyberMath]

www.wolframalpha.com/input?i2d=true&i=Power%5Bx%2Clogx%5D%3DPower%5B%5C%2840%29Sqrt%5Bx%5D%5C%2841%29%2Clnx%5D

[@yoav613]

@@SyberMath no syber,what you wrote is x^(lnx)=x^(lnx/2) because in wolfram logx is lnx.if you want logx in wolfram you should write log10(x)

[@XJWill1]

@@SyberMath It even printed a warning telling you that "assuming log is the natural logarithm". You should really read the warnings that WA gives. They can be helpful.

[@luker.6967]

I can't seem to convince Wolfram Alpha to give me the complex solutions for log base 10. However it would give them for log base 2 and 3, which were an absolute mess.

[@XJWill1]

That is because there are no complex solutions for log base 10. The solutions shown in the video are not for the equation given in the video. I'm not sure how he messed that up.

[@hichammouaddib7762]

😁👍

[@kianmath71]

X = 1

[@scottleung9587]

Interesting exploration of WA's answers!

[@ftorum19]

Equastion with complex numbers (sinx)^3+(cosx)^3=2

[@trojanleo123]

x = 1

[@michaelbaum6796]

Wow, this complex solution is very interesting 👍

[@yoav613]

Very interesting,but not asolution to this problem.

[@hydropage2855]

@@yoav613what are you talking about. It is a solution

[@alextang4688]

x^log(x)=(sqrt(x))^ln(x)
Take ln both sides,
ln(x^log(x))=ln((sqrt(x))^ln(x))
log(x)*ln(x)=ln(x)*ln(x^(1/2))
Therefore,
ln(x)=0 or log(x)=ln(x^(1/2))
case 1,
ln(x)=0
x=1 answer
case 2,
log(x)=ln(x^(1/2))
2log(x)=ln(x)
2log(x)=log(x)/log(e)
Therefore log(x)=0
x=1 answer
So, the answer is x=1 only. 😋😋😋😋😋😋

[@lazizbeknomozov7814]

f(x+y)+f(x-y)=2f(x)cosy f(x)=? x,y€R plz help me

[@antonyqueen6512]

You can check the result by using the formulas:
cos(x+y)=cos(x).cos(y) - sin(x).sin(y)
And
sin(x+y)=sin(x).cos(y) + cos(x).sin(y)

[@rakenzarnsworld2]

x = 1