Why does the same inequality give different answers? (log_2(x)−2)(log_2(x)+1)<0 has a solution 1.2<x<4 But when we take the second part alone that is (log_2(x)+1)<0 it gives a solution 0<x<1/2 why is x>1/2 in the first case but x<1/2 in the second case?

Why does the same inequality give different answers?
$\left({\mathrm{log}}_{2}\left(x\right)-2\right)\left({\mathrm{log}}_{2}\left(x\right)+1\right)<0$
has a solution $\frac{1}{2}
But when we take the second part alone that is
$\left({\mathrm{log}}_{2}\left(x\right)+1\right)<0$
it gives a solution $0 why is $x>\frac{1}{2}$ in the first case but $x<\frac{1}{2}$ in the second case?
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Cindy Walls
To solve this, you need exactly one of ${\mathrm{log}}_{2}\left(x\right)-2$ and ${\mathrm{log}}_{2}\left(x\right)+1$ to be negative. The problem reduces to solving the simultaneous inequalities:

and

to get two solution sets.
So to answer your question, you're correct that ${\mathrm{log}}_{2}\left(x\right)+1<0$ when $0, but in this region, we also have ${\mathrm{log}}_{2}\left(x\right)-2<0$, so overall the inequality is positive.