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Topic: x^2 = 2^x and x^4 = 4^x
Replies: 1   Last Post: Oct 28, 2013 11:40 AM

 Ben Bacarisse Posts: 1,972 Registered: 7/4/07
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 UTC-7, 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
> non-integer 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.