Exponential equation $(x^x)^{2015}=2015$

Question

Solve for $x$: $(x^x)^{2015}=2015$

Tried several times, but have no idea about how to start even.

Answer 1

Here's a start $$(1) \quad (x^x)^{2015}=2015$$ $$(2) \quad x^{2015 \cdot x}=2015$$ take the natural log of (2)... $$(3) \quad (2015 \cdot x) \cdot \ln(x)=\ln(2015)$$ $$(4) \quad \ln(x)={{\ln(2015)} \over {2015 \cdot x}}$$ finally, $$(5) \quad x=2015^{(2015 \cdot x)^{-1}}$$ What is this you might be wondering? Using the theory of dynamical systems we can see that the derivative of this function on the right side of (5) is less than 1. This means iterating (5) will approach the solution of (1)!

Thus the solution is...

$$x=2015^{\left(2015 \cdot 2015^{\left(2015 \cdot 2015^{\left(2015 \cdot 2015^{\left(2015....\right)^{-1}} \right)^{-1}} \right)^{-1}} \right)^{-1}}$$

This converges extremely fast. Just one convergent, $2015^{{1} \over {2015}}$, yields $x=1.00378...$. Very close indeed!

Answer 2

$$x^x=\sqrt[2015]{2015}\to x=e^{W(\ln(2015)/2015)}.$$