## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x\displaystyle \in\{\frac{1}{100,000},\quad 100,000\}$
In order for the equation to be defined, $\left\{\begin{array}{l} x\gt 0\\ x^{\log x}\gt 0 \end{array}\right.\qquad(*)$ apply $\quad\log_{a}M^{p}=p\cdot\log_{a}M$ $\log x\cdot\log x=25\qquad$ ... let $t=\log x$ $t^{2}=25$ $t=\pm 5$ $\left[\begin{array}{lll} \log x=-5 & ...or... & \log x=5\\ x=10^{-5} & & x=10^{5} \end{array}\right]$ Both satisfy (*), so they are both valid solutions $(10^{-5})^{\log 10^{-5}}=(10^{-5})^{-5}=10^{25}\gt 0$ $(10^{5})^{\log 10^{5}}=(10^{5})^{5}=10^{25}\gt 0$