Mathematic Problems
${x}^{1+{\mathrm{log}}_{10}\left(x\right)}=10x\phantom{\rule{0ex}{0ex}}$
${\mathrm{log}}_{10}{\left(x\right)}^{1+{\mathrm{log}}_{10}\left(x\right)}={\mathrm{log}}_{10}\left(10x\right)\phantom{\rule{0ex}{0ex}}$
$\left(1+{\mathrm{log}}_{10}\left(x\right)\right){\mathrm{log}}_{10}\left(x\right)={\mathrm{log}}_{10}\left(10x\right)\phantom{\rule{0ex}{0ex}}$
$\left(1+{\mathrm{log}}_{10}\left(x\right)\right){\mathrm{log}}_{10}\left(x\right)={\mathrm{log}}_{10}\left(10\right)+{\mathrm{log}}_{10}\left(x\right)\phantom{\rule{0ex}{0ex}}$
${\mathrm{log}}_{10}\left(x\right)=t\phantom{\rule{0ex}{0ex}}$
$\left(1+t\right)×t=1+t\phantom{\rule{0ex}{0ex}}$
$t+{t}^{2}=1+t\phantom{\rule{0ex}{0ex}}$
$t+{t}^{2}-1-t=0\phantom{\rule{0ex}{0ex}}$
${t}^{2}-1=0\phantom{\rule{0ex}{0ex}}$
$t=±\sqrt{1}\phantom{\rule{0ex}{0ex}}$
${\mathrm{log}}_{10}\left(x\right)=1\phantom{\rule{0ex}{0ex}}$
$x={10}^{1}⇒x=10\phantom{\rule{0ex}{0ex}}$
${\mathrm{log}}_{10}\left(x\right)=-1\phantom{\rule{0ex}{0ex}}$
$x={10}^{-1}⇒x=\frac{1}{10}\phantom{\rule{0ex}{0ex}}$
$x\in \left\{{10}^{1},{10}^{-1}\right\}$