Re: roots of x^12 = 2^x
- From: Phil Carmody <thefatphil_demunged@xxxxxxxxxxx>
- Date: 11 Apr 2007 16:26:51 +0300
"Ioannis" <morpheus@xxxxxxxxxxxx> writes:
"David W. Cantrell" <DWCantrell@xxxxxxxxxxx> wrote in message
news:20070410190915.721$e0@xxxxxxxxxxxxxxxxx
Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
In article <1176239376.759713@athprx04>,
"Ioannis" <morpheus@xxxxxxxxxxxx> wrote:
"chapkovski" <chapkovski@xxxxxxxxx> wrote in message
news:1176236640.212574.296720@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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)
Thanks to both for spotting the omission. I did an evalf(") on the exact
solutions Maple provided to scan for real solutions, but I missed that one,
which seems to be close to x_3 ~= 74.66932553
Moral of the story: When an equation is solvable exactly by W, *always* check
the -1 branch as well :-)
Not at all. The moral of the story is to not wave great hefty tools
like Lambert W functions around when techniques that the average
10 year old should know provide more insight into the solution.
I am aghast that you seemed to think that x^12 would dominate 2^x
as x increased.
Phil
--
"Home taping is killing big business profits. We left this side blank
so you can help." -- Dead Kennedys, written upon the B-side of tapes of
/In God We Trust, Inc./.
.
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