$$4^x - 7(2^{\frac{x-3}{2}}) = 2^{-x}$$ Set of real solutions is in which interval:
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?
On
$$4^x-7\left(2^{\frac{x-3}{2}}\right)=2^{-x}\Longleftrightarrow$$ $$2^{2x}-7\left(2^{\frac{x-3}{2}}\right)=2^{-x}\Longleftrightarrow$$ $$2^{3x}-7\left(2^{\frac{3x}{2}-\frac{3}{2}}\right)=1\Longleftrightarrow$$
Substitute $y=-7\left(2^{\frac{3x}{2}-\frac{3}{2}}\right)$:
$$\frac{8y^2}{49}+y=1\Longleftrightarrow$$ $$y^2+\frac{49y}{8}=\frac{49}{8}\Longleftrightarrow$$ $$y^2+\frac{49y}{8}+\frac{2401}{256}=\frac{3969}{256}\Longleftrightarrow$$ $$\left(y+\frac{49}{16}\right)^2=\frac{3969}{256}\Longleftrightarrow$$ $$y+\frac{49}{16}=\pm\sqrt{\frac{3969}{256}}\Longleftrightarrow$$ $$y+\frac{49}{16}=\pm\frac{63}{16}\Longleftrightarrow$$ $$y=\pm\frac{63}{16}-\frac{49}{16}\Longleftrightarrow$$
Notice that the positive one from our $y$ soution,
has no solution since for all $z\in\mathbb{R},2^z>0$ and $-\frac{1}{8}<0$:
$$y=-\frac{63}{16}-\frac{49}{16}\Longleftrightarrow$$ $$y=-7\Longleftrightarrow$$ $$-7\left(2^{\frac{3x}{2}-\frac{3}{2}}\right)=-7\Longleftrightarrow$$ $$2^{\frac{3x}{2}-\frac{3}{2}}=1\Longleftrightarrow$$ $$\frac{3x}{2}-\frac{3}{2}=0\Longleftrightarrow$$ $$\frac{3(x-1)}{2}=0\Longleftrightarrow$$ $$x-1=0\Longleftrightarrow$$ $$x=1$$
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 $$2^{\frac{3x}{2}}=\frac{7+9}{4\sqrt 2}=2\sqrt 2=2^{\frac{3}{2}}$$ $$\implies \frac{3x}{2}=\frac{3}{2}$$ $$\color{red}{x=1}$$