How to solve $x^x=100$?

Solution 1:

$$\begin{align} x^x&=100\\ x\log x&=\log 100\\ e^{\log x}\log x&=\log 100\\ \log x&=W(\log 100)\\ x&=e^{W(\log 100)}\\ x&\approx 3.597 \end{align}$$ Where $W(x)$ is the ProductLog function, defined as the inverse of $[f(x)=xe^x]$

Explanations: 1. Problem

2.Logs

3.$ e^{\log x}=x$

4.Definition of $W(x)$

5.Taking $e^x$

6.Numerical Solution

Solution 2:

$$3^3=27\,\,\,\,\,\,\,\,\,\,\,\,\,\ 4^4=256$$ Therefore there are no integer solutions for this equation. The unique solution of this equation with $50$ decimal places can obtain using Mathematica as

$$3.5972850235404175054976522517822860691355430548866.$$

Also I plot the graph of $f(x)=x^x-100$ in $x\in[3,4].$

Solution 3:

x=log(100)/log(x)

x'=x/2+log(100)/2log(x)

After several iteration for x (that is x', x'' to xn) you'll come to a good value of x to so much decimal places.

What am doing up there is applying the Babylonian method (an iterative algorithm similar to Newton-Raphson method). I've reduced it to its empirical form. Self taught I should add.