Question

If ${\log }_{3/4}\left({\log }_8\right(x^2+7\left)\right)+{\log }_{1/2}\left({\log }_{1/4}\right(x^2+7{)}^{-1})=-2$, then sum of all values of $x$ is equal to

Answer 1

${\log }_{3/4}{\log }_8(x^2+7)+{\log }_{1/2}{\log }_{1/4}(x^2+7{)}^{-1}=-2$ .....(1)
Let ${\log }_8(x^2+7)=a$
$\Rightarrow x^2+7=8^a$ ....(2)
Also let ${\log }_{1/4}(x^2+7{)}^{-1}=b$
$\Rightarrow -{\log }_{1/4}(x^2+7)=b$
$\Rightarrow {\log }_4(x^2+7)=b$
$\Rightarrow x^2+7=4^b$ ....(3)
From eqn (2) and (3), we get
$8^a=4^b$
$\Rightarrow 2^{3a}=2^{2b}$
$\Rightarrow 3a=2b$
$\Rightarrow b=\frac{3a}{2}$
Now, by eqn (1),
${\log }_{3/4}a+{\log }_{1/2}\left(\frac{3a}{2}\right)=-2$
${\log }_{3/4}a+{\log }_{1/2}\left(3a\right)-{\log }_{1/2}2=-2$
${\log }_{3/4}a+{\log }_{1/2}\left(3a\right)+{\log }_{2}2=-2$
${\log }_{3/4}a+{\log }_{1/2}\left(3a\right)=-3$ ....(4)
Now, let ${\log }_{3/4}a=p$
$\Rightarrow a=(\frac{3}{4}{)}^p$ ....(5)
Also, let ${\log }_{1/2}\left(3a\right)=q$
$\Rightarrow 3a=(\frac{1}{2}{)}^q$
$\Rightarrow a=\frac{1}{3}(\frac{1}{2}{)}^q$ ....(6)
So eqn (4) can be written as
$p+q=-3$
$\Rightarrow p=-(q+3)$
From eqn (5) and (6),
$(\frac{3}{4}{)}^p=\frac{1}{3}(\frac{1}{2}{)}^q$
$\Rightarrow (\frac{3}{4}{)}^{-(q+3)}=\frac{1}{3}(\frac{1}{2}{)}^q$
$\Rightarrow 2^{3q+6}=3^{q+2}$
$\Rightarrow 8^{q+2}=3^{q+2}$
which is possible only when $q+2=0$
Hence, $q=-2$
$\Rightarrow p=-1$
So, by eqn (5)
$a=\frac{4}{3}$
So,by eqn (2),
$x^2+7=8^{\frac{4}{3}}$
$\Rightarrow x^2+7=16$
$\Rightarrow x^2=9$
$\Rightarrow x=\pm 3$