# How to solve log(x)+2x=2 for x?

• B
cakeroot
Hello:
could anyone know how to solve this equation?
log(x)+2x+2=0
it can be solve by lambert w function,
but in this equation we can't reshape it to xe^x that I can't solve by lambert w function.

Thanks!

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Orodruin
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Yes you can apply the W function to this. It is perfectly possible to rewrite your equation in terms of it.

Apart from that, you will not find a closed form solution.

hilbert2
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You can guess that x=1, but for some other numbers like log(x)+1.5x=4 you have to solve it iteratively because it's a transcendental equation. Lambert W is just a name that has been given to the results of that kind of iterations.

Orodruin
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Lambert W is just a name that has been given to the results of that kind of iterations.
Like sin and cos are also just names that we have given other transcendental functions. Just to point out that it is not something strange and magical about the W.

cakeroot
Yes you can apply the W function to this. It is perfectly possible to rewrite your equation in terms of it.

Apart from that, you will not find a closed form solution.

but I still encountered difficulties in this problem,

the following is my derivation process：

##log(x)=-2x-2##

##x = e^{-2-2x}##

##xe^{2+2x}=1##

##xe^{2x} = \frac{1}{e^2}##

how can I solve it by lambert w?

or others solutions?

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hilbert2
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Like sin and cos are also just names that we have given other transcendental functions. Just to point out that it is not something strange and magical about the W.

Yes, I should probably have stressed that fact more. The W function may be said to be a bit more exotic object that sine or cosine, though, as its power series and the differential equation producing it as solution are not as simple as those of sin and cos. Also, it doesn't appear in any fundamental physical model (sin and cos are found in the energy eigenfunctions of a free particle and the trajectory of harmonic oscillator).

mfb
Mentor
xe^2x = 1/(e^2)

how can I solve it by lambert w?
By definition, ##W(z)e^{W(z)}=z##. Here ##z=\frac{1}{e^2}##, and your solution is W(z).

Edit: Forgot the factor 2. See two posts below.

What exactly is the initial problem statement? Title and first post have two different equations.

Last edited:
cakeroot
cakeroot
By definition, ##W(z)e^{W(z)}=z##. Here ##z=\frac{1}{e^2}##, and your solution is W(z).

What exactly is the initial problem statement? Title and first post have two different equations.

I apologize for my negligence,

the first post is right => log(x)+2x+2=0,

---------------------------------------------------
the left side of equation is ##xe^{2x}##,
and if shift to ##xe^{x}## will cause:
##xe^{x}=\frac{1}{e^{2+x}}##,
then the right side included x that I can't solve it by w function.

please correct me if I am wrong,
thanks

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Orodruin
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##xe^{2x} = \frac{1}{e^2}##

how can I solve it by lambert w?

or others solutions?

Multiply both sides by 2 ........

Douglas Sunday and cakeroot
cakeroot
By definition, ##W(z)e^{W(z)}=z##. Here ##z=\frac{1}{e^2}##, and your solution is W(z).

What exactly is the initial problem statement? Title and first post have two different equations.
Multiply both sides by 2 ........

ah, ok, I think I understand what to do now,
thanks!