$(2^x-8)(x-7)(x+1) \leq 0)$ to $(x-3)(x-7)(x+1)\leq 0$

Question

$(2^x-8)(x-7)(x+1) \leq 0 $
in solutions author change $(2^x-8)$ to $(x-3) $
and solve it as $(x-3)(x-7)(x+1)\leq 0$
how he do this?
can you provide another solution?

Answer 1

The sign of the product depends only on the signs of the factors. Hence it is perfectly valid to replace $2^x-8$ with $x-3$. Both are positive iff $x>3$, negative iff $x<3$ and zero iff $x=3$.

Answer 2

Because $\log_2(x)$ is strictly monotonic $2^x \gt 8$ is equivalent to $x \gt 3$ so the signs of $2^x-8$ and $x-3$ are the same.