Trying to solve a math equation, arg.

Passions

Diamond Member
Feb 17, 2000
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I need to solve this with precision:

x^x=0.606530659

When I put it into my HP 48G, it solves

x=0.367879479

But, when you put that back in to check, x=0.692200647

ARGGGGGGGG.....???
 

spp

Golden Member
Jul 9, 2001
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Originally posted by: spp
hmm can't you take natural log of that and solve it?

oops... yea use natural log and then use euler's method to keep approximating it?? you should be able to get to pretty precise answer
 

Brian23

Banned
Dec 28, 1999
1,655
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if x^2=.606530659 and
e^(-.5)=.606530659 then
x^2=e^(-.5)
x=sqrt(x^(-.5))
and there you go!


EDIT -- oops! I read the equation wrong.

according to my TI 89, the CORRECT answer is this:

x*ln(x)=-.5
I dunno if that helps
 

spp

Golden Member
Jul 9, 2001
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Originally posted by: Brian23
if x^2=.606530659 and
e^(-.5)=.606530659 then
x^2=e^(-.5)
x=sqrt(x^(-.5))
and there you go!

dude.... it's x^x not x^2......
 

KMurphy

Golden Member
May 16, 2000
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ans =

.32447650802884661101302231982777+.31470495693373390251468816640960*i