<s>[INST] A Question on which solution is false and where (with logarithms) This question was put to me yesterday during a meeting with one of my highschool school teacher who taught me mathematics.
The question:
Find the domain of the function $f:x \to$ $log_\frac{1}{10}  \frac{2x+1}{x+2}$.From this, solve the inequality $f(x) \ge log_\frac{1}{10}  \frac{1}{x}$.
Solution for the domain of the function:
Since $ \frac{2x+1}{x+2} = \frac{2x+4-3}{x+2}$ = 2- $\frac{3}{x+2} \ge 0$,then $x \in $$(-\infty,-2) \cup (-\frac{1}{2},\infty):=S$
Here comes the funny part, my teacher offered two solutions for solving the inequality:
solution 1 (which is also similar to mine) :
With the premise of $x \in S = D_f$, we have that
$log_\frac{1}{10}  \frac{2x+1}{x+2} \ge log_\frac{1}{10}\frac{1}{x}$
$\iff$ $\frac{2x+1}{x+2} \le \frac{1}{x}$ $\land$ $\frac{1}{x} \gt 0$
$\iff$ $\frac{2(x+1)(x-1)}{x(x+2)} \le 0$ $\land$ $x \gt 0$
$\iff$ $x-1 \le 0 \land x \gt 0$
$\iff$ $x \le 1 \land x \gt 0$
Hence $x \in (0,1]$
Solution 2(which I personally think that there's something wrong with this solution) :
Consider $f(x) \ge log_\frac{1}{10}  \frac{1}{x}$
Choose y such that $f(y) = log_\frac{1}{10}  \frac{1}{x}$ $\land$ $x \gt 0$
so
$\begin{cases} f(x) \ge f(y)\\log_\frac{1}{10} (2-\frac{3}{y+2})=log_\frac{1}{10}\frac{1}{x}\\ x \gt 0 \land y\in S\end{cases}$
$\begin{cases} x \le y  ,&\text{(because $f$ is decreasing)} \\ 2-\frac{3}{y+2}=\frac{1}{x} ,&\text{(take the $antilog={0.1}^{x}$ of both sides)} \\ x \gt 0 \land y\in S\end{cases}$
$\begin{cases} x+2\le y+2\\ y+2 =\frac{3x}{2x-1}\\ x \gt 0 \land y\in S\end{cases}$
sub the second equation to the first inequality and we will get
$x+2 \le \frac{3x}{2x-1} \land x,2x-1 \gt 0$
$(x+2)(2x-1)-3x \le 0 \land x\gt \frac{1}{2} $
$2(x-1)(x+1) \lt 0  \land x\gt \frac{1}{2} $
$(x-1)\lt 0  \land x\gt \frac{1}{2} $
Hence $ \frac{1}{2} \lt x \le 1 $
Which solution is wrong and where?
 [/INST] The second solution is wrong because $f$ is only decreasing on each of the two intervals $(-\infty,-2)$ and $(-\frac12,\infty)$ but not on their union.
So, a large forgotten part of the solution is when $y<-2$ (and $x>0$, and $y=\frac{2-x}{2x-1}$), i.e. $0<x<\frac12.$
Moreover, the value $x=\frac12$ is also forgotten because there is no $y$ such that $\frac{2y+1}{y+2}=2.$
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