# Exercise ID333

Algebra → Logarithm → Logarithm equations
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Evaluate the following expression:

${x}^{1+{\mathrm{log}}_{10}\left(x\right)}=10x\phantom{\rule{0ex}{0ex}}$

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