# Exercise ID330

Algebra → Logarithm → Logarithm equations
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 18] [Number of characters: 0]

Evaluate the following expression:

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

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