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Re: x^2 = 2^x and x^4 = 4^x
Posted:
Oct 28, 2013 11:40 AM


Ray Vickson <RGVickson@shaw.ca> writes:
> On Monday, October 28, 2013 6:30:46 AM UTC7, James Ward wrote: >> I didn't see this curious fact mentioned in the wiki article (perhaps >> it is well known) http://en.wikipedia.org/wiki/Tetration >> >> The two equations: >> >> x^2 = 2^x and x^4 = 4^x, >> >> both have 3 identical real solutions: >> >> x = 2, 4, and infinite power tower of (1/sqrt(2)) >> >> You can check the last in Wolfram Alpha using: >> >> x = 1/sqrt(2), >> y = ProductLog((log(x)))/(log(x)), >> z = 2^y  y^2 >> >> and >> >> x = 1/sqrt(2), >> y = ProductLog((log(x)))/(log(x)), >> z = 4^y  y^4 > > It follows by simple algebra: if x^2 = 2^x, then (x^2)^2 = (2^x)^2, so > x^4 = (2^2)^x = 4^x. In fact, for any positive integer n we have that > x^2 = 2^x implies x^(2n) = (2^n)^x. It is even true for positive > noninteger values of n. No need for Wolfram Alpha here.
I don't think that's the "curious" bit  I suspect it's the third solution that uses Lambert's W function (which is related to tetration) that James considers curious.
 Ben.



