You might solve this math exam question wrong, be careful

[@wes9627]

I have never used the Lambert W() function because it is not on my calculator, but I have iteratively solved for all three real roots of this and similar equations, either by using fixed-point or Newton-raphson iteration. Everyone should become proficient in using these iterative methods to solve a wide range of nonlinear systems of equations.

[@zawatsky]

Эволюция математика: от "да это устно решается!" до "неужели нет способа найти все корни без функции Ламберта?!".🤭

[@Checkyourself.Physics.Math-GB]

Similar problem solved without Lambert function: kzbin.info/www/bejne/hHjGY6Z6l9CYaZY

[@slickback_was_taken]

This is mental math. 😂😂

[@easystat1929]

Why divideb by 2x it's confusing

[@CanerDeniz1]

[@rainerzufall42]

(x > 0!) - ln x / x = - ln x / e^(ln x) = - ln x e^(- ln x) => W_n(- ln x / x) = - ln x, as long als ln x / x < 1 / e (for W_0 branch). Now, x=2 is positive and ln 2 / 2 is 0.69314... / 2 = 0.34657... and 1 / e = 0.36788..., so ln 2 / 2 < 1 / e. Thus follows W(- ln 2 / 2) = - ln 2 = -0.69314... and x_1 = 1 / e^W_0(-ln 2 / 2) = e^(ln 2) = 2. So much for the W_0 branch! ln |x| / x = ln 2 / 2 => W_k(- ln |x| e^(- ln |x|)) = W_l(- ln 2 e^(- ln 2)) = W_m(- ln 4 e^(- ln 4)), l = 0. Probably m=-1, have to check. The problem starts with x=4. There are reasons, why W_0 is not sufficient here! One is that ln 2 / 2 = 2 ln 2 / 4 = ln 2² / 4 = ln 4 / 4. There is no difference in ln x / x for x=2 and x=4. Thus you need another branch to explicitly select the solution. For x < 0, it's even more complicated: ln x is not available, nor is x = e^(ln x) true. You must circumvent this with a special case. As you said, something with ln |x|. To my own shame, I saw x=2 and x=4, but I oversaw the negative solution (that is only possible because of the even exponent!)... shit happens!

[@rainerzufall42]

Why didn't you do the same thing on the right side - immediately? (I expect, you'll do it later in the video).

[@rainerzufall42]

Okay, you didn't. My answer has moved to the comment root...

[@ManojkantSamal]

X=2 because if a^m=m^a, then a=m Another method X^2=2^x Take the log logx^2=log 2^x 2.log x=x. log2 logx/x=log2/2 So, X=2

[@ainamanicolleb]

logx/x=log2/2 how did you get that?

[@ManojkantSamal]

@ainamanicolleb , Taking the antilog of the equation....

[@боженкогеоргий]

4?

[@SEBE3835]

Better watch the video because you only found 1 solution of the 3.

[@ManojkantSamal]

Nice, A better way to shape this problem as of Lambert w function...

[@RyanLewis-Johnson-wq6xs]

X=2 X=4 X^2=2^X

[@therozer1203]

How x=e^ln x ?? I didn't get it

[@l0l_h8d_l0l5]

The e and the ln cancel out

[@payoo_2674]

e^x is the inverse function of ln(x) ln(x) is the inverse function of e^x so e^ln(x) = x and ln(e^x) = x The same principle applies to Lambert's W function, which is the inverse function of xe^x so W(xe^x) = x

[@payoo_2674]

x₁ = e^(-W(-ln(2)/2)) = e^(-W(-ln(2)*(2^(-1))) = e^(-W(-ln(2)*e^ln(2^(-1))) = e^(-W(-ln(2)*e^(-ln(2))) = e^(-(-ln(2))) = e^ln(2) = 2 x₂ = e^(-W(-ln(2)/2)) = e^(-W(-(2*ln(2))/(2*2))) = e^(-W(-ln(2^2)/4)) = e^(-W(-ln(4)*(4^(-1))) = e^(-W(-ln(4)*e^ln(4^(-1))) = = e^(-W(-ln(4)*e^(-ln(4))) = e^(-(-ln(4))) = e^ln(4) = 4 x₃ = -0.766664695962123093111204422510314848006675346669832058460884376...

[@johnbernardmakiling3991]

I just did the trial-and-error way, it's way simpler

[@presentnaol1364]

its only work if you have simple answer as he show here for exemple but in real situation, it is way more complicated

[@zawatsky]

По какому алгоритму эта загадочная функция считается, так никто в итоге и не объяснил. %)

[@giannaleoci2328]

x=2

[@frankcollins4743]

It took me less than TWO seconds to work out the answer can only be TWO. Nothing else works.

[@rainerzufall42]

So 4² isn't equal to 2^4 ?

[@RealQinnMalloryu4]

(x ➖ 2x+2). {2x+2x ➖}=4x^2 2^2x^2 1^1x^2 1x^2(x ➖ 2x+1).