# Exercise ID331

Algebra → Logarithm → Logarithm equations
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Evaluate the following expression:

${\mathrm{log}}_{x}\left(3\right)+{\mathrm{log}}_{3}\left(x\right)={\mathrm{log}}_{\sqrt{x}}\left(3\right)+{\mathrm{log}}_{3}\left(\sqrt{x}\right)+\frac{1}{2}\phantom{\rule{0ex}{0ex}}$

$\frac{1}{{\mathrm{log}}_{3}\left(x\right)}+{\mathrm{log}}_{3}\left(x\right)=\frac{1}{{\mathrm{log}}_{3}\left({x}^{\frac{1}{2}}\right)}+{\mathrm{log}}_{3}\left({x}^{\frac{1}{2}}\right)+\frac{1}{2}\phantom{\rule{0ex}{0ex}}$ $\frac{1}{{\mathrm{log}}_{3}\left(x\right)}+{\mathrm{log}}_{3}\left(x\right)=\frac{1}{\frac{1}{2}{\mathrm{log}}_{3}\left(x\right)}+\frac{1}{2}{\mathrm{log}}_{3}\left(x\right)+\frac{1}{2}\phantom{\rule{0ex}{0ex}}$ ${\mathrm{log}}_{3}\left(x\right)=t\phantom{\rule{0ex}{0ex}}$ $2+2{t}^{2}=4-{t}^{2}+t\phantom{\rule{0ex}{0ex}}$ $2+2{t}^{2}-4+{t}^{2}-t=0\phantom{\rule{0ex}{0ex}}$ ${t}^{2}-t-2=0\phantom{\rule{0ex}{0ex}}$ ${t}_{1,2}=\frac{1±\sqrt{1+8}}{2}\phantom{\rule{0ex}{0ex}}$ ${t}_{1,2}=\frac{1±3}{2}\phantom{\rule{0ex}{0ex}}$ ${\mathrm{log}}_{3}\left(x\right)=2\phantom{\rule{0ex}{0ex}}$ $x={3}^{2}⇒x=9\phantom{\rule{0ex}{0ex}}$ ${\mathrm{log}}_{3}\left(x\right)=-1\phantom{\rule{0ex}{0ex}}$ $x={3}^{-1}⇒x=\frac{1}{3}\phantom{\rule{0ex}{0ex}}$
$x={3}^{-1}⇒x=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$