2 different exponential equations

We will solve 2 different exponential equations: x^2*e^x=2 and x+e^x=2 by using the Lambert W function.

Solving equations with Lambert W functions,

blackpenredpen

[pwootjuhs]

x+e^x=2
e^x=2-x
x=ln(2-x)
x=ln(2-ln(2-ln(2-ln(...))))

[carlbianzon]

I tried differently for example 2... my instinct was to raise e to both sides to get rid of the "+"
e^(x+e^x) = e^2
(e^x)[e^(e^x)] = e^2
e^x = W(e^2)
x = ln[W(e^2)]
which is also around 0.44285...

BTW nice video! I'm new to this concept as it wasn't taught (not even mentioned) in school! Good thing YouTube recommended this for me!

[fCauneau]

On the second one, let y=e^x, this gives shortly the solution :
Ln(y)+y = 2
Ln(y)+y.Ln(e) = 2
Ln(y)+Ln(e^y)=2
y e^y=e^2
y=W(e^2)
x=Ln(W(e^2))

[xaxuser]

《Who put the square. .... yeh that person was me》
《OMG, who put the + he was me 》
😂legend

[alexs]

bprp: "okay lets do some math for fun"
me: *I N S T A N T R A G I N G B O N E R*

[Koisheep]

Even now I'm graduated in Math I'm still learning from you lmao

[PackSciences]

As I've read the comment, noone has been talking about that; the method used in equation 2 can be used to demonstrate the Wien's displacement law from the Planck law.
You try to find the maximum of the Planck law as a single variable (lambda), with T being a fixed parameter; then you should end up with (x-5)e^x + 5 = 0.
And it's exactly the same method as BPRP used in this video, but I am going to do it for the physics students if they want a detailed answer:
(x-5)e^x + 5 = 0
e^5(x-5)e^(x-5) = - 5
(x-5)e^(x-5) = -5e^-5
Apply W in both sides:
x-5 = W(-5e^-5)
x = 5 + W(-5*e^-5)

[pietpadda]

Great video. I never learned about the Lambert W function for my engineering degree and was always fascinated by the problems my friends and I considered "unsolvable". Your videos helped a lot!

[szmatogowiec]

That's pretty cool, *isn't it?* :D

[night]

It'd be cool if there was a page of equations like these ones to solve.

[doctor]

I got a similar answer for the second one. I used the substitution x = ln(y), and I plugged both sides into the exponential function, which got me the equation y*e^y = e^2. Applying the lambert function, and reversing the substitution, this got me ln(W(e^2)) = x.

[shohamsen]

For problem 2, wouldn't it be easier to eponentiate both sides. You get e^x e^(e^x)=e^2 => ye^y=e^2.=> x=ln[W(e^2)]. Where we are using y=e^x.

[lucarl]

For the 2nd one :

x + e^x = 2
exp(x + e^x) = exp(2)
e^x • exp(e^x) = e^2
W(e^x • exp(e^x)) = W(e^2)
e^x = W(e^2)
x = ln(W(e^2))

and just to match your answer :

W(e^x) = y
e^x = y • e^y
exp(x-y) = y
y = exp(x - W(e^x))

So,
ln(W(e^2)) = ln(exp(2 - W(e^2)) = 2 - W(e^2)

[alexismiller]

I found a weird answer for the second one:
x+e^x=2
e^(x+e^x)=e^2
e^x*e^(e^x)=e^2
with lambert W we get
e^x=W(e^2)
so x=ln(W(e^2))

Does that mean ln(W(e^2))=2-W(e^2)? I checked on Wolfram and yeah, they're the same:
This gives us a new identity:

ln(W(e^x))=x-W(e^x)

or in other words ln(W(x))=lnx-W(x)

Unsurprisingly, from this we can prove this identity easily since
ln(W(x))=lnx-W(x)
ln(W(x))+W(x)=lnx
so left hand side gives us
= ln(W(x))+ln(e^W(x))
= ln(W(x)*e^W(x))
= ln x

Proved (:

[hassanalihusseini]

Thank you blackpenredpen for making these videos about Lambert's W-function. Very nice explaining and creative solutions!

[slowfreq]

It seems a little arbitrary. What's stopping you from creating a function which is the inverse of whichever expression you're working with, and then saying that's the answer?

[Debg]

You have convinced me, I'll start using the Lambert function whenever I can 😉

[KirillBon]

I have seen several videos on your channel, but this one is the best !

[TheRealSamSpedding]

Mate you've become the Bob Ross of maths

[Rtong]

I actually tried to do it!! Good video
Also, do you think you could do a video on how to approximate the W function or at least how the computers do it? Thanks