Question

Solve the following equations.
$lo{g}_{1/3}\sqrt{x^2-2x}=sin\frac{11\pi }{6}$

Answer 1

$lo{g}_{\frac{1}{3}}^{\sqrt{x^2-2x}}=sin\left(\frac{11\pi }{6}\right)$

$sin\left(\frac{11\pi }{6}\right)=-\frac{1}{2}$

$\therefore lo{g}_{\frac{1}{3}}^{\sqrt{x^2-2x}}=\frac{-1}{2}$

$\Rightarrow \sqrt{x^2-2x}=\sqrt{3}$

$\Rightarrow x^2-2x=3$

$\Rightarrow x^2-2x-3=0$

$\Rightarrow (x+1)(x-3)=0$

$x=-1,3$

$x=-1$ is not a valid solution,

log can't take negative solve.

Answer :3