# Do we flip the inequality symbol when dividing or multiplying by an expression variable?

Do we flip the inequality symbol when dividing or multiplying by an expression variable the same way we do when we multiply with or divide by a negative number?

I am currently having some confusion about understanding logarithmic inequalities.

There is one inequality that I’m not sure why I got incorrect.

$$\log\left(\frac{2x-1}{x-2}\right) / \log2 < 0$$

1. For it to be real, I know that $$(2x-1)/(x-2) > 0,$$ and so we should be able to cancel out (multiply both sides) the expression variable $$(x-2).$$ Without changing the symbol from $$>$$ to $$<$$ after multiplication however, I got the result $$x > 1/2$$ instead of the correct answer $$x < 1/2.$$

2. I also know that $$(2x-1)/x-2 < 2^0,$$ and so we should also be able to cancel out the expression variable $$(x-2).$$ Without changing the symbol from $$<$$ to $$>$$ after multiplication however, I got the result $$x < -1$$ instead of the correct answer $$x > -1.$$

After combining both cases in a number line, the correct answer is $$-1 < x < 1/2,$$

but I got $$x < -1, x > 1/2,$$ except $$x=2,$$ which is tested to be incorrect.

I’m not sure if the not switching of symbols is the source of my error, which is why I am asking.

Because this process involves multiplying or dividing expression variables, I also feel that my method is not suitable due to the possibility of extraneous solutions, and maybe should draw and fill in a trial+error table instead.

• you cannot flip the inequality unless and until you're absolutely sure about the nature of the variable(positive/negative, $>1,<1$ etc) Aug 19, 2020 at 8:12
• 1. should be $(2x-1)/(x-2)>1$. Aug 19, 2020 at 8:17
• Thing is, you do not know if $x-2$ is negative or positive. If it is negative, it will flip the inequality sign. The trick is, multiply by $(x-2)^{2}$ instead because square is non negative Aug 19, 2020 at 8:18
• noted, thank you for the comments. Aug 19, 2020 at 8:40

We have $$\frac{2x-1}{x-2}>0$$. Multiplying by $$(x-2)^2$$ on both sides, we have

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

and hence $$x < \frac12$$ or $$x > 2$$.

Your mistake is that you have assume that $$x-2>0$$ must be true.

We also know that $$\frac{2x-1}{x-2}<1$$

If $$x>2$$, then we have $$2x-1 < x-2$$, which is equivalent to $$x < -1$$ which contradicts $$x>2$$.

If $$x < \frac12$$, then we have $$2x-1 > x-2$$ and hence $$x > -1$$.

The conclusion is $$-1 < x < \frac12$$.

• Thank you very much for your answer, but I made a mistake in typing the question, which should be log... <0 and not log... >0, which was why your answer did not match. I have recently edited it. So sorry about that. Nonetheless, I now realized my mistake and got the correct answer. Aug 19, 2020 at 8:34
• I have modified the post according to your edit. Aug 19, 2020 at 8:41