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I'm trying to solve this inequation: $\ln\left(\dfrac{ 2x}{ x-1}\right) > 0$. The problem is, $2$ different ways that seem valid to me, give me different answers.

First option: turn $0$ to $\ln(1)$, so we have $\ln\left(\dfrac{ 2x}{x-1}\right) > \ln(1) \to \dfrac{2x}{x-1} > 1$ and multiply both sides by $(x-1)^2$ gives me $x > 1$ or $x < -1$.

Second option: $\ln\left(\dfrac{ 2x}{x-1}\right)$ is actually $\ln(2x) - \ln(x-1)$, so $\ln(2x) - \ln(x-1) > 0 \Rightarrow \ln(2x) > \ln(x-1)$ so $2x > x-1$, or \Rightarrow x > -1$.

I'm guessing the second option is the invalid one, but why?

Thanks!

amWhy
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Stars
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1 Answers1

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The first method is correct and the second one is wrong.

At first note that we require

$$\frac{2x}{x-1}>0 \iff x<0 \quad \land \quad x>1$$

then we obtain

$$\log\left(\frac{2x}{x-1}\right)>\log 1\iff \frac{2x}{x-1}>1 \iff \frac{2x}{x-1}-1>0\iff \frac{x+1}{x-1}>0$$

that is

$$x\in(-\infty, -1)\cup(1,\infty)$$

which is the solution.

The second method leads to a inequality which is not equivalent since

$$\log(2x)-\log(x-1)>0$$

requires that

$$x>0\quad \land \quad x-1>0 \iff x>1$$

The reason for the second method to fail is that

$$\log \frac A B= \log A-\log B$$

only holds when $A$ and $B$ are both positive.

To obtain a correct solution, we can also use the second method considering all the cases as follows

  • for $x<0$

$$\log\left(\frac{2x}{x-1}\right)>0 \iff \log(-2x)-\log(1-x)>0$$

  • for $0<x<1$

$$\log\left(\frac{2x}{x-1}\right)>0 \iff \log(2x)-\log(1-x)>0$$

  • for $x>1$

$$\log\left(\frac{2x}{x-1}\right)>0 \iff \log(2x)-\log(x-1)>0$$

which is a less effective way to proceed.

user
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  • I am not sure it is clear how you reached the conclusion in the last line from the one before. Thanks. – NoChance Nov 11 '19 at 23:16
  • just consider the sign for numerator and denominator – user Nov 11 '19 at 23:18
  • That’s for the existence of log expression. – user Nov 11 '19 at 23:20
  • Thank you for your answer, I need to spend sometime to understand this one. – NoChance Nov 12 '19 at 00:34
  • Thanks for your answer. I too don't see how the last line is a conclusion of the line before it. Nevertheless, we agree that the two options are not equivalent and give different answers. I'm wondering how could that be, since all the actions were legit and according to logarithms rules. I think the real question here is how could it be that for y=ln(2x/x-1), x=-10 (for example) is valid, but for y=ln(2x)-ln(x-1), which supposed to be the same according to logarithms rules, x=-10 is invalid. – Stars Nov 12 '19 at 08:01
  • For the ratio $\frac{x+1}{x-1}$ we have that the numerator is positive for $x>1$ and negative for $x<1$ while the denominator is positive for $x>-1$ and negative for $x<-1$ therefore the ratio is positive for $x<-1$ or $x>1$. – user Nov 12 '19 at 08:10
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    @Stars I’ve added some more detail to explain why the second method leads to a different inequality. In general observe that $\log \frac A B= \log A-\log B$ only holds when $A$ and $B$ are both positive. That’s the origin of the issue with the second method. – user Nov 12 '19 at 08:14
  • @user Thanks for the detailed response! – Stars Nov 12 '19 at 09:14
  • Thanks for the detail. – NoChance Nov 12 '19 at 09:49
  • @Stars You are welcome! Bye – user Nov 12 '19 at 10:22
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    @NoChance You are welcome! Bye – user Nov 12 '19 at 10:22