Question

Solve the following equations.
$lo{g}_2(3-x)-lo{g}_2\frac{sin\frac{3\pi }{4}}{5-x}=\frac{1}{2}+lo{g}_2(x+7)$

Answer 1

$lo{g}_2(3-x)-lo{g}_2\left(\frac{sin\frac{3x}{4}}{5-x}\right)=\frac{1}{2}+lo{g}_2(x+y)$

We know that

$lo{g}^{mn}=lo{g}^{\left(m\right)}+lo{g}^{\left(n\right)}...\left(i\right)$

$lo{g}^{\left(m/n\right)}=,lo{g}^{\left(m\right)}-lo{g}^{\left(n\right)}...\left(ii\right)$

Now,

$lo{g}_2^{(3-x)}-lo{g}_2^{\left(\frac{sin\frac{3\pi }{4}}{5-x}\right)}-lo{g}_2^{(x+y)}=\frac{1}{2}$

Using properties (i) & (ii)

$lo{g}_2\frac{(3-x)(5-x)}{(x+7)sin\left(\frac{3x}{4}\right)}=\frac{1}{2}$

$\Rightarrow \sqrt{2}\frac{(3-x)(5-x)}{(x+7)}=\sqrt{2}$

$(\because sin\frac{3\pi }{4}=\frac{1}{\sqrt{2}})$

$\Rightarrow (3-x)(5-x)=(x+7)$

$\Rightarrow x^2-9x+8=0$

$x=1,8$ ( 8 is not valid)

Checking the solution by

putting in original equation,

$x=8$ is not valid solution

(log can't have negative value)

$\therefore$ Answer $:1$