<s>[INST] Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter $a$. Question : Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$   for every value of the parameter a.
I have solved the problem as follows
$\sqrt{a\left(2^{x}-2\right)+1}=1-2^{x}$
$a\left(2^{x}-2\right)+1=\left(1-2^{x}\right)^{2}=2^{2 x}+1-2 \cdot 2^{x}=2^{2 x}-2^{x+1}+1$
$a 2^{x}-2 a=2^{2 x}-2^{x+1}$
$2^{2 x}-(a-1) 2^{x}+2 a=0$
$y^{2}-(a-1) y+2 a=0$
$y=\frac{(a-1) \pm \sqrt{(a-1)^{2}-8 a}}{2}=\frac{(a-1) \pm \sqrt{a^{2}-10 a+1}}{2}$
$2^{x}=\frac{(a-1) \pm \sqrt{(a-5)^{2}-24}}{2}$
After this there are so many conditions on a. Do i need to check for each and every value ?
 [/INST] $$\sqrt{a(2^{x}-2)+1}=1-2^{x}\tag1$$
First of all, we have to have
$$a(2^{x}-2)+1\ge 0\qquad\text{and}\qquad 1-2^x\ge 0\tag2$$
Under $(2)$, we have
$$\begin{align}(1)&\implies a(2^x-2)+1=(1-2^x)^2
\\\\&\implies a(2^x-2)+1=1-2^{x+1}+2^{2x}
\\\\&\implies a\cdot 2^x-2a=2^{2x}-2\cdot 2^{x}
\\\\&\implies 2^{2x}+2^x(-2-a)+2a=0
\\\\&\implies (2^x-2)(2^x-a)=0
\\\\&\implies 2^x=2\quad \text{or}\quad 2^x=a\end{align}$$
Now, $2^x=2$ does not satisfy $(2)$.
When $2^x=a$, we have to have $0\lt a\le 1$ from $(2)$.
So, the answer is as follows :


*

*If $a\le 0$ or $a\gt 1$, then there is no $x$ satisfying $(1)$.

*If $0\lt a\le 1$, then $x=\log_2 a$.
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