Solving $\ln(\frac{ 2x}{ x-1}) > 0$

Question

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!

Answer 1

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.