Re: roots of x^12 = 2^x
- From: Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Tue, 10 Apr 2007 22:48:27 GMT
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)
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.
--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.
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