Solve the inequality x times log base 1/3 of (1/2 - x) is greater than or equal to the absolute value of x.

Step-by-step solution

1. 1

   Fix the domain before touching the inequality. A logarithm needs a strictly positive argument:

   $\frac12-x>0\quad\Longleftrightarrow\quad x<\frac12$

   Everything below is searched inside $\left(-\infty,\tfrac12\right)$; any candidate outside it is discarded automatically.

2. 2

   Split on the sign of $x$, because both $|x|$ and the direction of the inequality depend on it. Write $L=\log_{1/3}\!\left(\tfrac12-x\right)$. The inequality is $xL\ge|x|$. Dividing by $x$ is the natural move, but the sign of $x$ decides both what $|x|$ becomes and whether the inequality flips, so the three cases $x>0$, $x=0$, $x<0$ must be handled separately.

3. 3

   Case $x>0$. Here $|x|=x$, and dividing by the positive number $x$ preserves the direction:

   $L\ge 1$

   The base $\tfrac13$ is less than $1$, so $\log_{1/3}$ is decreasing: $\log_{1/3}t\ge 1$ reverses into $0<t\le\left(\tfrac13\right)^{1}=\tfrac13$. Hence

   $0<\frac12-x\le\frac13\quad\Longleftrightarrow\quad \frac16\le x<\frac12$

   Combined with $x>0$ this gives $\left[\tfrac16,\tfrac12\right)$.

4. 4

   Case $x<0$. Now $|x|=-x$, and dividing $xL\ge-x$ by the negative number $x$ flips the inequality:

   $L\le -1$

   Again using that $\log_{1/3}$ is decreasing, $\log_{1/3}t\le-1$ becomes $t\ge\left(\tfrac13\right)^{-1}=3$:

   $\frac12-x\ge3\quad\Longleftrightarrow\quad x\le-\frac52$

   Every such $x$ is negative and inside the domain, so the whole ray $\left(-\infty,-\tfrac52\right]$ qualifies.

5. 5

   Case $x=0$ — do not skip it. Substituting directly gives $0\cdot L=0$ and $|0|=0$, so the inequality reads $0\ge0$, which is true. The isolated point $x=0$ is a genuine solution even though it sits in neither interval; it is exactly the point that a careless division by $x$ would have lost.

6. 6

   Assemble and spot-check. The solution set is

   $x\in\left(-\infty,-\frac52\right]\cup\{0\}\cup\left[\frac16,\frac12\right)$

   Check the two boundary points: at $x=-\tfrac52$, $\tfrac12-x=3$ and $\log_{1/3}3=-1$, so the left side is $(-\tfrac52)(-1)=\tfrac52=|x|$ — equality holds. At $x=\tfrac16$, $\tfrac12-x=\tfrac13$ and $\log_{1/3}\tfrac13=1$, so the left side is $\tfrac16=|x|$ — equality again. A test value in a rejected zone, say $x=-1$, gives $(-1)\log_{1/3}\tfrac32\approx0.369$, which is less than $|x|=1$, so it is correctly excluded.