Question

Solve the equation ${\log }_{3/4}{\log }_8(x^2+7)+{\log }_{1/2}{\log }_{1/4}(x^2+7{)}^{-1}=-2$ and find the value of $x$

Answer 1

${\log }_{3/4}{\log }_8(x^2+7)+{\log }_{1/2}{\log }_{1/4}(x^2+7{)}^{-1}=-2$

$\Rightarrow \frac{{\log }_2{\log }_{2^3}(x^2+7)}{{\log }_{2}3/4}-{\log }_2{\log }_{2^{-2}}(x^2+7{)}^{-1}=-2$

$\Rightarrow \frac{{\log }_2\left(\frac{1}{3}{\log }_2\right(x^2+7))}{{\log }_{2}3/4}-{\log }_2\left(\frac{1}{2}{\log }_2(x^2+7)\right)=\frac{{\log }_2{\log }_2(x^2+7)-{\log }_{2}3}{{\log }_{2}3/4}-{\log }_2{\log }_2(x^2+7)+1=-2$

Let ${\log }_2{\log }_2(x^2+7)=y$

$\Rightarrow y(\frac{1}{{\log }_{2}3/4}-1)=\frac{{\log }_{2}3}{{\log }_{2}3/4}-3={\log }_{3/4}(3\times (4/3{)}^3)$

$\Rightarrow y\left({\log }_{3/4}2\times \left(4/3\right)\right)=y\left({\log }_{3/4}8/3\right)={\log }_{3/4}(8/3{)}^2$

$\Rightarrow y=2\Rightarrow {\log }_2{\log }_2(x^2+7)=2$

$\Rightarrow {\log }_2(x^2+7)=4\Rightarrow (x^2+7)=2^4$

$\Rightarrow x^2=9\Rightarrow x=\pm 3$