Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter $a$.

Question

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 ?

Answer 1

By letting $y=1-2^x$, you can simplify the problem to: $$ \sqrt{a(3-y)+1} = y $$

Since root is non-negative, $y\ge0$. Since $2^x$ is always positive, $y<1$. Thus we want to find the solution of the equation above with the condition $0\le y<1$.

Since both sides are non-negative, we can square them and keep the equality: $$ a(3-y)+1=y^2\\ y^2+ay - (1+3a)=0 $$

This equation will have the solution when $D=a^2+4(1+3a)\ge 0$. By solving the corresponding equation, we will get $a\in I_1=(-\infty;-6-4\sqrt2]\cup[-6+4\sqrt2;+\infty)$.

Solution for $y$: $$ y_{1,2}=\frac12\left(-a\pm\sqrt{a^2+12a+4}\right). $$

Now let's check when $y\ge0$. For smaller root: $$ y_1\ge0\\ -a-\sqrt{a^2+12a+4} \ge 0\\ -a \ge \sqrt{a^2+12a+4}\\ a\le0\qquad\text{and}\qquad a^2\ge a^2+12a+4 $$

So, smaller root is always smaller than 0. For larger root: $$ y_2\ge 0\\ -a+\sqrt{a^2+12a+4} \ge 0 \\ \sqrt{a^2+12a+4}\ge a\\ \text{either}\qquad a<0\qquad \text{or }\qquad a^2+12a+4\ge a^2 \\ \text{either}\qquad a<0\qquad \text{or }\qquad a\ge -1/3 $$

Thus, $y_2\ge 0$ when exists.

Finally, we need to check that $y<1$. We don't need to check the $y_1$, since it already failed previous condition. For larger root: $$ y_1 < 1\\ -a+\sqrt{a^2+12a+4} < 2\\ \sqrt{a^2+12a+4} < 2+a\\ a\ge-2\qquad\text{and}\qquad a^2+12a+4 < (2+a)^2 \\ a\ge-2\qquad\text{and}\qquad 12a+4 < 4a+4 \\ a\ge-2\qquad\text{and}\qquad a< 0 \\ $$

Thus, $a\in I_1 \cap[-2; 0)=[-6+4\sqrt2; 0)$. Finally, the answer:

When $a\in [-6+4\sqrt2; 0)$, the equation has one root, which is: $$ x = \log_2\left(1-\frac{\sqrt{a^2+12a+4}-a}2\right). $$

Note. I didn't include analysis of that expression under the root should be non-negative when was calculating $y\ge0$ and $y<1$, because we already did this analysis, and the corresponding interval would be intersecting the result anyway.

Note 2. We didn't need to check that $a(3-y)+1\ge 0$, because we set it equal to $y^2$ which is always non-negative

Answer 2

$$\sqrt{a(2^{x}-2)+1}=1-2^{x}\tag1$$

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 $x=\log_2 a$.

Answer 3

With $z:=2^x-2$, $$\sqrt{az+1}=-z-1$$

or after rewriting,

$$z(z+2-a)=0,z\le-1.$$

Hence

$$z=a-2\le-1,$$ which is $$2^x=a\le1$$

or

$$x=\log_2a,\\0<a\le1.$$