The number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$

Question

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?

Answer 1

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 $0<2^x <1$.

Answer 2

There are two cases:

1. $x\geq0$. 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 $x = 2$.

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 $x=2$.

Answer 3

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 $$ a = 4 \implies 2^x = 2 \implies \boxed{x = 2}. $$

Answer 4

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.