<s>[INST] The set of real solutions to an equation $$4^x - 7(2^{\frac{x-3}{2}}) = 2^{-x}$$
Set of real solutions is in which interval:


*

*$(-9, -2)$

*$(0, 3]$

*$(-2, 0]$

*$(7, 12]$

*$(3, 7]$


I tried the following. Dividing by $2^{-x}$ I get $2^{3x} - \frac{7\sqrt(2)}{4}2^{\frac{3x}{2}} - 1 = 0$. Substituing 
$$
2^{3x} = c
$$ and solving I get $c_1= 2\sqrt(2)$ and $c_2=-\frac{\sqrt(2)}{4}$ and for $x$ I get $\frac{1}{2}$. That's not even in the given line. What is wrong with the calculations. What have I missed?
 [/INST] Notice, $$4^x-7\cdot 2^{\frac{x-3}{2}}=2^{-x}$$
$$(2\sqrt2)2^{3x}-7\cdot 2^{\frac{3x}{2}}-2\sqrt2=0$$
let $2^{\frac{3x}{2}}=t$, 
$$(2\sqrt 2)t^2-7t-2\sqrt 2=0$$
$$t=\frac{-(-7)\pm\sqrt{(-7)^2-4(2\sqrt 2)(-2\sqrt 2)}}{2(2\sqrt2)}=\frac{7\pm9}{4\sqrt 2}$$ but $t(=2^{\frac{3x}{2}})>0$ hence, one should get 
$$2^{\frac{3x}{2}}=\frac{7+9}{4\sqrt 2}=2\sqrt 2=2^{\frac{3}{2}}$$
$$\implies \frac{3x}{2}=\frac{3}{2}$$ $$\color{red}{x=1}$$
</s>