Question

Solve the equation
${\log }_2(3-x)-{\log }_2\left(\frac{sin\frac{3\pi }{4}}{5-x}\right)=\frac{1}{2}+{\log }_2(x+7)$

Answer 1

The given equation is equivalent to
${\log }_2(3-x)={\log }_2\left(\frac{\sin \frac{3\pi }{4}}{5-x}\right)+\frac{1}{2}{\log }_{2}2+{\log }_2(x+7)$
$\Rightarrow {\log }_2(3-x)={\log }_2\left(\frac{1}{\sqrt{2}(5-x)}\right)+{\log }_2\sqrt{2}+{\log }_2(x+7)$
which is equivalent to the system
$\{\begin{array}{c}3-x>0\\ \frac{1}{\sqrt{2}(5-x)}>0\\ x+7>0\\ (3-x)=\frac{\sqrt{2}(x+7)}{\sqrt{2}(5-x)}\end{array}\Rightarrow \{\begin{array}{c}x<3\\ x<5\\ x>-7\\ (x-1)(x-8)=0\end{array}$
Hence $x=1$ is the only root of the original equation.