2
$\begingroup$

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.

Thank you in advance for your time and answers.

$\endgroup$
4
  • $\begingroup$ you cannot flip the inequality unless and until you're absolutely sure about the nature of the variable(positive/negative, $>1,<1$ etc) $\endgroup$
    – sai-kartik
    Aug 19, 2020 at 8:12
  • $\begingroup$ 1. should be $(2x-1)/(x-2)>1$. $\endgroup$
    – user375366
    Aug 19, 2020 at 8:17
  • $\begingroup$ 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 $\endgroup$ Aug 19, 2020 at 8:18
  • $\begingroup$ noted, thank you for the comments. $\endgroup$
    – user74261
    Aug 19, 2020 at 8:40

1 Answer 1

1
$\begingroup$

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 $-1 < x < \frac12$.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – user74261
    Aug 19, 2020 at 8:34
  • $\begingroup$ I have modified the post according to your edit. $\endgroup$ Aug 19, 2020 at 8:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .