Finding the unknown variable

Question

What is the value of $x$ in $x^{x}=25$? How can this be solved in the easiest way of all? I just couldn't deduce any idea regarding where to start.

Answer 1

\begin{align} x^x &= 25 \\ x^2 &= 25 \end{align} possibilities, $x=5$ or $x=-5$ because $5^2 = 25$ and $-5^2 =25$ description: \begin{align} -5\times-5 &=(-1)\times5\times(-1)\times5 \\ &=(-1)\times(-1)\times5\times5 \\ &=1\times5\times5 \\ &=25 \end{align}

Answer 2

The solution of $ x^x=a $ is $$ x = \frac{\log a}{W(\log a)} $$ where $W$ is the Lambert W function, not an elementary function.

Answer 3

$$ \begin{align} x^x&=25\tag{1}\\ x\log(x)&=\log(25)\tag{2}\\ \log(x)e^{\log(x)}&=\log(25)\tag{3}\\ \log(x)&=\mathrm{W}(\log(25))\tag{4}\\ x&=e^{\mathrm{W}(\log(25))}\tag{5}\\ &=\frac{\log(25)}{\mathrm{W}(\log(25))}\tag{6} \end{align} $$ Explanation:
$(2)$: take log of both sides
$(3)$: $x=e^{\log(x)}$
$(4)$: if $we^w=x$, then $w=\mathrm{W}(x)$, $\mathrm{W}$ is Lambert W
$(5)$: exponentiate $(4)$
$(6)$: divide $(2)$ by $(4)$