Axe Power Axe!

From $\space x^x = 55 \quad \Rightarrow x\ln x = \ln 55$ .

By Newton's method, we have: $\space x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$ .

Let $f(x) = x\ln x - \ln 55\quad \Rightarrow f'(x) = \ln x + 1$ .

Using the following Excel spreadsheet, we found that $x = \boxed{3.330651551}$ .

Because Brilliant's floating point answer only cares about three significant digits, the fact that the problem asks for 6 decimal places is irrelevant. I simply brute-forced it. We know the answer is somewhere between $3$ and $4$ (because $3^3 = 27 < 55 < 256 = 4^4$ ), and that $x \mapsto x^x$ is strictly increasing for $x \ge 1$ , so:

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for i in range(300, 400): # 3.00 <= i/100 <= 3.99
    if 55 < (i/100) ** (i/100): # the first i/100 to cross 55
        print(i/100)
        break
# prints 3.34

That's why, whenever you want to force computation to a specific precision, you ask for something like $\lfloor 10^\text{precision} \cdot \text{answer} \rfloor$ instead. A better method that doesn't abuse Brilliant's way of handling floating point answers is given by Chew-Seong Cheong.

Using the Python sympy module:

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>>> from sympy import *
>>> var('x')
x
>>> solve(x**x-55)
[exp(LambertW(log(55)))]
>>> N(_[0])
3.33065155117834

Using the Python scipy module:

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from scipy.optimize import bisect

f=lambda x:x**x-55

print bisect(f,3,4)

The solution with the C Programming. For this problem, n=55.

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#include<stdio.h>
#include<math.h>  
/*As we used the POWER function*/
int main(){
double n,x=2;
/*The Equation is X^x=n
So let's solve it.
X^x=n
or, X=n^[1/x]
So we should take different values of X sequentially and get n^[1/x].
When the LHS and RHS are equal, its the solution!
*/
printf("Stuck on Solving X^X=n, right?\nJust provide the n, I'll detect the value of X.\nAnd have patience!!\n So, n=");
scanf("%lf",&n);
while(1){
if((pow(n,(1/x))-x)<=pow(10,-15)){
printf("X=%.10lf\n",x);
break;
}
x=x+pow(10,-6); //Lower the power of 10 for higher precision! But remember, it'll take more time, maybe enough to lose your patience!

}
return 0;
}

And the program gives $\boxed{3.330652……}$ . Remind that, its a Transcendental Number . So you can't get the actual, but you have an approximation!