What are the products of real solutions of this equation? How can I solve $\:\: \log^2_{1/2}(4x)+\log_2\hspace{-0.06 in}\left(\hspace{-0.06 in}\frac{x^2}{8}\hspace{-0.06 in}\right)=8 \;$ ?
I have tried the elementary for logarithms simplifying the terms in brackets.
 A: Let $\log_2 x = t$.  Then 
$$\log_{1/2} (4x) = \frac{\log_2(4x)}{\log_2(1/2)} = -(2+t)$$
$$\log_2\frac{x^2}{8} = 2t-3$$
So we solve $(2+t)^2+2t-3=8 \iff (t-1)(t+7)=0$, or $x=2, \dfrac1{2^7}$ for a product of $\frac1{64}$.
A: Going to natural logarithms (the only I know, if I may confess), you have  $$\log_{1/2}(4x)=-\frac{\log (4 x)}{\log (2)}=-2-\frac{\log ( x)}{\log (2)}$$ $$\log_2(\frac{x^2}{8})=\frac{\log \left(\frac{x^2}{8}\right)}{\log (2)}=\frac{2\log ( x)}{\log (2)}-3$$ So, setting $t=\frac{\log ( x)}{\log (2)}=\log_2(x)$, $$\log^2_{1/2}(4x)+\log_2(\frac{x^2}{8})=(2+t)^2+2t-3$$ and after development, the equation to solve is then $$t^2+6t-7=(t-1)(t+7)=0$$
A: $\left( { { log }_{ 2^{ -1 } } }\left( 4x \right)  \right) ^{ 2 }+{ 2log }_{ 2 }x-{ log }_{ 2 }2^{ 3 }=8\quad $ $\Rightarrow $ $\left( { log }_{ 2 }4+{ log }_{ 2 }x \right) ^{ 2 }+2{ log }_{ 2 }x-3=8$  $ \Rightarrow $ $\left( 2+{ log }_{ 2 }x \right) ^{ 2 }+2{ log }_{ 2 }x=11$  $ \Rightarrow $ $4+6{ log }_{ 2 }x+{ log }_{ 2 }^{ 2 }x=11$ $ \Rightarrow $ ${ log }_{ 2 }^{ 2 }x+6{ log }_{ 2 }x-7=0$
let call $${ log }_{ 2 }x=t$$ then your equation will be $$t^{ 2 }+6t-7=0$$
which you can proceed 
