roots of x^12 = 2^x - sci.math

Message from discussion roots of x^12 = 2^x

David W. Cantrell

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More options Apr 10 2007, 7:09 pm

Newsgroups: sci.math

From: David W. Cantrell <DWCantr...@sigmaxi.net>

Date: 10 Apr 2007 23:09:12 GMT

Local: Tues, Apr 10 2007 7:09 pm

Subject: Re: roots of x^12 = 2^x

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Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
> In article <1176239376.759713@athprx04>,
> "Ioannis" <morph...@olympus.mons> wrote:

> > "chapkovski" <chapkov...@gmail.com> wrote in message
> > news:1176236640.212574.296720@a30g2000cwd.googlegroups.com...

> > > how many roots does this equation have?

> > Two real ones, approximately at x_0 ~= -.9467803304 and at x_1 ~=
> > 1.063346831, given by Lambert's W function as:

> > x = -12*W((+/-) log(2)/12)/log(2)

> 2^x is smaller than x^{12} at x = -1,
> bigger at x = 0,
> smaller at x = 2,
> and bigger at x = 84 ( 2^{84} = (2^7)^{12} = 128^{12} > 84^{12} ),
> so three real solutions.

Yes indeed. The two roots given by Ioannis are obtained using the principal
branch of the Lambert W function. The third root is given by

x = -12*W_{-1}(-log(2)/12)/log(2)

where W_{-1} denotes the -1 branch.

David

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