Question

Solve for $x$, ${\log }_2\left(2\sqrt{17-2x}\right)=1-{\log }_{1/2}(x-1)$

Answer 1

${\log }_2\left(2\sqrt{17-2x}\right)=1-{\log }_{1/2}(x-1)$ is valid when $17-2x>0,x>1$
$\Rightarrow x\in (1,\frac{17}{2})$
Rewrite as $1+{\log }_2\left(\sqrt{17-2x}\right)=1+{\log }_2(x-1)$
$\Rightarrow \sqrt{17-2x}=(x-1)$
$\Rightarrow 17-2x=x^2-2x+1$
$\Rightarrow x=\pm 4$
$x>1\Rightarrow x=4$