The equation was not completely solved. There is one more solution. Since the number (-ln4)/8 belongs to the interval (-1/e , 0) the Lambert function W_-1 must also be applied which gives the solution x = 0.15495 For more details and the complete solution of the equation x^x=a see my channel named L+M=N at: www.youtube.com/@Nikos_Iosifidis/videos The solution of the equation x^x =a is at: th-cam.com/video/f8jmmnfu7wA/w-d-xo.html I'd love for you to subscribe and comment on the topics I present.
is there a way to algebraically apply the W_-1 function? for example, say I wanted to take W(-ln4/4), -ln/4 is in the interval (-1/e , 0), so after simplifying W_0(-ln4/4) like this: W_0(4^-1*ln4^-1) -> W_0(e^(ln(4^-1))*ln4^-1) -> ln4^-1 -> -ln4. Is there a way to do a similar simplification with the W_-1 branch?
0.155 is another root, but this can be found out by Desmos graph. How to Know this other root as 0.155 ? Do you know how to arrive at 0.155 without using graph ?
La ecuación tiene dos soluciones, como han dicho varios: 2 y 0.15495 (aprox). Las resoluciones gráficas con posterior aproximación de raices es aquí muy apropiado. Y evita el uso erróneo de transformaciones, funciones y complejos programas que no siempre se saben manejar.. Un estudio básico de la función 2^2x-8x (o directamente grafico con geogebra o cualquier programa) ubica los intervalos de sus raices, que son las de la ecuación. Luego de hacer esto si no me convencí que hay dos raices.... Lo último, no le veo aporte ninguno a la función de Lambert. Alguno llega a creer que sólo con ella se puede resolver esta ecuación.
For a value of x equal to 2, we have a solution, but it is not the solution based on Lambert's W function presented here. For the given argument, the value of x equals 0.1550, which is confirmed by implementations in MATLAB, Python. How did you get the value of 2 ?
If you have two variables but only one equation, you can set the value of a to anything you like and then solve for b. There are infinitely many solutions.
Every Video falls short of evaluating the Lambert W Function because no one has tabulated the numerical values of this mysterious, useless Lambert W function, Don't waste your time on this bit of trickery.
Great video
To be precise, the value of the argument of Lambert's W function is equal to -ln(4)/8=0.1733, the value of W=-0.2148, and x=-W/ln(4)=0.1550
I also calculated the LambertW function result using WolframAlpha's Productlog function, and I got the same answer.
See my comment
👍👍👍
The equation was not completely solved. There is one more solution. Since the number (-ln4)/8 belongs to the interval (-1/e , 0) the Lambert function W_-1 must also be applied which gives the solution x = 0.15495
For more details and the complete solution of the equation x^x=a see my channel named L+M=N at: www.youtube.com/@Nikos_Iosifidis/videos
The solution of the equation x^x =a is at: th-cam.com/video/f8jmmnfu7wA/w-d-xo.html
I'd love for you to subscribe and comment on the topics I present.
is there a way to algebraically apply the W_-1 function? for example, say I wanted to take W(-ln4/4), -ln/4 is in the interval (-1/e , 0), so after simplifying W_0(-ln4/4) like this: W_0(4^-1*ln4^-1) -> W_0(e^(ln(4^-1))*ln4^-1) -> ln4^-1 -> -ln4. Is there a way to do a similar simplification with the W_-1 branch?
and the other value ? ~ 0.155
0.155 is another root, but this can be found out by Desmos graph.
How to Know this other root as 0.155 ?
Do you know how to arrive at 0.155 without using graph ?
@@nitingl4730 Lambert function
Only the Lambert W function(wolfram alpha)
La ecuación tiene dos soluciones, como han dicho varios: 2 y 0.15495 (aprox). Las resoluciones gráficas con posterior aproximación de raices es aquí muy apropiado. Y evita el uso erróneo de transformaciones, funciones y complejos
programas que no siempre se saben manejar.. Un estudio básico de la función 2^2x-8x (o directamente grafico con geogebra o cualquier programa) ubica los intervalos de sus raices, que son las de la ecuación. Luego de hacer esto si no me convencí que hay dos raices....
Lo último, no le veo aporte ninguno a la función de Lambert. Alguno llega a creer que sólo con ella se puede resolver esta ecuación.
Can this equation be solved taking logarithm on both sides ?
Not at all
Try and see what happens... No succes. @@petrochengula9480
直接心算2
For a value of x equal to 2, we have a solution, but it is not the solution based on Lambert's W function presented here. For the given argument, the value of x equals 0.1550, which is confirmed by implementations in MATLAB, Python. How did you get the value of 2 ?
See my comment
Using Wolfram software I didn't get "2" for x, but 0.155...
There are 2 solutions: x=2 and x=0.155...
X=2
8×2=16
16=2^(2×2)
16=2⁴
16=16
if a√5 + b√2 is the square root of 95 - 30√10, the values of "a" and "b" are respectively __
Pls someone solve
If you have two variables but only one equation, you can set the value of a to anything you like and then solve for b.
There are infinitely many solutions.
搞了那麼多根本看不懂的鬼東西，還不如我的直接心算x=2
Every Video falls short of evaluating the Lambert W Function because no one has tabulated the numerical values of this mysterious, useless Lambert W function, Don't waste your time on this bit of trickery.
Can be able to get help in partial differential equations