Solve:$x^x=5$ and $x^{(1/x)}=1/5$ The first equation comes from standard log equation:
$\log(x)= \frac{1}{x}$
When base of log is $5$ then it will be reduced to
$x^x=5$
The second equation comes from standard log equation:
$\log(x)=-x$
When base of log is $5$ then it will be reduced to
$x^{1/x}= \frac{1}{5}$
 A: The first equation is
$$x^x=5$$
First, take the natural log of both sides:
$$x\log x=\log5$$
$$\implies e^{\log x} \log x = \log 5$$
Now, take the product log (Lambert W function) of both sides:
$$W(e^{\log x} \log x)=W(\log 5)$$
Which gives
$$\log x=W(\log 5)$$
i.e.,
$$x=e^{W(\log 5)}$$
Similarly, the second can be solved. The solution of the second is $e^{-W(\log5)}$.
A: As @Leonhard Euler answered, the only analytical solutions involve Lambert function.
If, for any reason, you cannot use it, you will need some numerical method (Newton being the simplest) but, as usual, you will need to provide a starting guess.
For the first one, consider that you look for the zero('s) of function
$$f(x)=x^x-5$$ It is varying very fast. So, assuming $x>0$, consider instead the much smoother function
$$g(x)=x\log(x)-\log(5)$$ Its derivatives are
$$g'(x)=\log(x)+1 \qquad \text{and} \qquad g''(x)=\frac 1 x \quad >0 \quad \forall x >0$$
The first derivative cancels at
$$x_*=\frac 1e \implies g(x_*)=-\frac{1}{e}-\log (5) <0$$ and the second derivative test confirms that $x_*$ correspond to a minimum.
Around this point, build a Taylor series to have as an estimate
$$g(x)=g(x_*)+\frac 12 g''(x_*)(x-x_*)^2  +O\left((x-x_*)^3\right)$$ Ignoring the higher order terms, solving the abova equation
$$x_0=x_*\pm \sqrt{-2\frac{g(x_*)}{g''(x_*)}}$$
$$x_0=\frac{1}{e}\pm\frac{\sqrt{2 (1+e \log (5))}}{e}$$ The negative solution must be dicarded because we assumed $x>0$ and our guess is then
$$x_0=\frac{1}{e}+\frac{\sqrt{2 (1+e \log (5))}}{e}\approx 1.57404$$ while the exact solution is $2.12937$. Now, using Newton, the iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 1.5740423 \\
 1 & 2.1899953 \\
 2 & 2.1298472 \\
 3 & 2.1293725
\end{array}
\right)$$
For the second one, in the same spirit
$$f(x)=x^{\frac 1x}-\frac 15$$
$$g(x)=\log(x)+x\log(5)$$
$$g'(x)=\frac{1}{x}+\log (5) \qquad \text{and} \qquad g''(x)=-\frac{1}{x^2}\quad <0 \quad \forall x >0$$
The first derivative cancels at
$$x_*=-\frac{1}{\log (5)} \implies g(x_*)=-\frac{1}{e}-\log (5) <0$$ and the second derivative test confirms that $x_*$ correspond to a minimum.
Just continue the same process.
