X^x=100 findx? these question is still like a black magic to me

Answer 1

this has no algebraic solution. You need some kind of graphical solution. Take a look at the graph of

x^x - 100 = 0

and you will see that x ≈ 3.6

http://www.wolframalpha.com/input/?i=x^x-100+for+3%3Cx%3C4

That is reasonable, since
3^3 = 27
4^4 = 256

Answer 2

But steve my math teacher told me to apply somekind newtons law in it don't no how any ideal??

Answer 3

Newton's Method perhaps?

x^x = 100
take ln of both sides, and use log rules
xlnx = ln100

let y = f(x) = xlnx - ln100
dy/dx = x(1/x) + lnx + 0
= 1 + lnx

Newton is effect said:
newx = oldx - f(oldx)/f ' (oldx) , using x for oldx
= x - (xlnx - ln100)/(1+lnx)
= (x + xlnx - xlnx + ln100)/(1+lnx)
= (x + ln100)/(1+lnx)

picking a good starting x:
I know 3^3 = 27 and 4^4 = 256
so x is between 3 and 4 , (see Steve's solution)
I will start with x = 3

x , newx
3, 3.6239...
3.6239.. , 3.597...
3.597.. , 3.597285024
3.597285024 , 3.597285023
3.597285023 , 3.597285024
The answer bounces back between those two values

so x = 3.597285023..

check:
3.597285023^3.597285023 = 99.99999992 , not bad eh?

Answer 4

Your a genius tanks reiny