The number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$ I'm trying to find the number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$.
Let $2^x=a$
$$|a-1|=a^2-2a-5$$
Then there are two cases 
$$a-1=a^2-2a-5$$
And $$a-1=-a^2+2a+5$$
Solving both equations 
$$a=1,-4,-2,3$$
Now -4 and -2 can be neglected so there are two values 
1 and 3.
Then $$2^x=1$$
$$x=0$$
And $$2^x=3$$
$$x=\log_2 3$$
But the answer doesn’t seem to consider the $\log_2 3$ as a viable root, and the answer is 1. Why is that the case?
 A: 
Then there are two cases 
  $$a-1=a^2-2a-5$$
$$a-1=-a^2+2a+5$$
Solving both equations 
  $$a=1,-4,-2,3$$

You made a small mistake in solving these.
The first equation is $a^2 - 3a - 4 = 0$, so $(a - 4)(a + 1) = 0$, so $a = 4$ or $a = -1$.
Looks like you solved the second equation correctly. You should have
$$
a = -1, 4, -2, 3
$$

But the answer doesn’t seem to consider the $\log_2 3$ as a viable root

Once we have found the values of $a$ that work, we're not done yet!
We have to plug in the values to the original equation to see if they work. The problem is that $|a - 1| = a - 1$ only if $a \ge 1$, and $|a - 1| = -(a - 1)$ only if $a \le 1$, so we might have introduced some "extraneous" answers that are not valid.
As you observed, since $a = 2^x$, $a$ has to be positive; so we are left with just $a = 3$ and $a = 4$ as possibilities.
Next, we have to plug in each to the original equation to see if they work.
$a = 4$ works, but $a = 3$ doesn't. So the only answer is
$$
a = 4 \implies 2^x = 2 \implies \boxed{x = 2}.
$$
A: Your way is right indeed for $2^x \ge1$
$$5+|2^x-1|=2^x(2^x-2)\implies 5+2^x-1=2^{2x}-2\cdot 2^{x} \iff2^{2x}-3 \cdot 2^x-4=0$$
for $0<2^x <1$
$$5+|2^x-1|=2^x(2^x-2)\implies 5-2^x+1=2^{2x}-2\cdot 2^{x} \iff2^{2x}- 2^x-6=0$$
then let $2^x=t>0$ and solve keeping only the solutions which agree with the assumptions.
Note that $2^x=3$ is not valid since it was obtained under the assumption that  $0<2^x <1$.
A: There are two cases:


*

*$x\geq0$. The equation becomes $$2^{2x}-2^{x+1}-2^x-4=0$$$$(2^x-2)(2^x-1)= 6$$Clearly, $2^x-2>0$, which implies $x>1$. Let $y=2^x$. We have $y^2-3y-4=0$ which implies $$(y-4)(y+1)=0$$ So, $y=4$ and $x = 2$.

*$x<0$. The equation becomes $$6-2^x=2^{2x}-2^{x+1}$$$$2^{2x}-2^{x+1}+2^x-6=0$$$$(2^x-2)(2^x+1)=4$$Let $y=2^x$. Hence, $$y^2-y-6=0\to(y-3)(y+2)=0$$Neither solution can equal $2^x$ where $x<0$. So there are no solutions.
So, the only solution is $x=2$.
A: It's helpful to think in terms of solution sets. After the substitution $a=2^x$, what you found is that
$$|a-1|=a^2-2a-5\implies a\in\{1,-4,-2,3\}$$
The take-home lesson here is that there is a difference between saying
this and saying
$$|a-1|=a^2-2a-5\iff a\in\{1,-4,-2,3\}$$
But you also made another error: the solutions to the equation $a-1=a^2-2a-5$ aren't $1$ and $-4$, but $-1$ and $4$.  In fact what's true is
$$|a-1|=a^2-2a-5\iff a\in\{4,-2\}$$
after which the solution $a=-2$ is ruled out by the requirement that $a=2^x$ must be positive, leaving $2^x=4$, or $x=2$ as the only solution.
