On Jun 24, 9:20 pm, Alan Bromborsky <abro...@verizon.net> wrote:
> The equation reduces to exp(w) = 2*exp(2)*w where w = x*(x-3)+2.
> Wouldn't the solution of w have to be numerical?
> Is this what you are looking for?
...I think it's exp(w)=2*(w+2). And that's the key. When put in this
form, the online Mathematica at QuickMath knew what to do. And it
appears that the page Ondrej referred to also knew what to do without
substituting w for x(x-3).
I prefer to work all lambert-able equations down to the form
a*w+log(c*w+d)=b which has the solution w=LambertW(a*exp(b + a*d/c)/c)/
a - d/c.
What I wanted to see is whether Mathematica was recognizing that w can
be any solvable expression of x; if you can solve w=f(x)=K then you
can get a solution to the original equation. For example,
###
>>> solv(log(w)+2*w-3,w)
[LambertW(2*exp(3))/2]
>>> solv(x*exp(x)-3-_[0],x) #so w=x*exp(x)-3
[LambertW(3 + LambertW(2*exp(3))/2)]
>>> _[0].evalf()
1.24834688674721
###
And that is the numerical answer (given without symbolic form) at the
Mathematica site. So apparently some forms for "w" are recognized, but
not others.
/c
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