What is an analytic solution to x^x = 100?

Answer 1

Using Newton's method:

f(x) = x^x - 100 = 0

df/dx = x^x (1 + ln(x))

x(i+1) = x(i) - f(x(i)) / df(x(i))/dx

x = x - (x^x - 100) / (x^x (1 + ln(x))

x = x - (1 - 100 / (x^x)) / (1 + ln(x))

Iterate the last expression, it converges quadratically. After you get the first
two digits correct, each iteration gets twice as many good digits. The answer I
got using VPCalc was:

X = 3.59728,50235,40417,50549,76522,51782,28606,91355,43054,88657,67837,20252,12
797,29575,07555,97393,18180,35224,03705,91783,30204,97674,84 E+0

Answer 2

This can be easily solved in Microsoft excel by using the goal seek function. First we subtract 100 from both sides so that x^x-100 =0, Then we iterate until a solution of x is found which satisfies this equation. If you have Excel go to the tools section, scroll down until you see goal seek, enter the values and hey presto you solution is found to be 3.59728546397371.

Answer 3

There is no analytic way to solve.With a scientific calculator I
x=3.59728502354