I have a question about solving inequality logarithm with absolute x in it (attached in image). And i want to know if my work is right or not. Thanks.

enter image description here

  • $\begingroup$ Please have a look at the introduction to posting mathematical expressions. You will find there links to more advanced topics. $\endgroup$
    – hardmath
    Apr 8 '17 at 6:02
  • $\begingroup$ I think if you're asking for help it is respectful to put your work in a legible format. It's expected here that you will learn Latex and type maths in that. $\endgroup$ Apr 8 '17 at 7:14
  • $\begingroup$ Sorry. I can't put mathematic symbol here. i don't know how. $\endgroup$ Apr 8 '17 at 7:30

Assuming you want to show for what $x, \ log_{(1-|x|)}|(3x-1)|<1$ (with base 1-|x|), we have the following:

What is the inverse of $log|x|$? (i.e. its exponential form)

We also assume $logx$ has the base $e$ (when the base is not specified)

So upon inverting the logarithm to it's exponential equivalent, we have:

$$log_{(1-|x|)}|(3x-1)|<1\Rightarrow (1-|x|)<(3x-1)$$

Noting $logx$ is only defined for $x>0$, we thus immediately see x is bounded above by $1$. That is:

$$log(1-|x|) \Rightarrow x<1$$ Why? Consider $log(1-x)>0$

A lower bound for x follows from Considering the above inequality $(1-|x|)<(3x-1)$

Thus we have:

For $x>0$, $(1-x)<(3x-1)\Rightarrow 2<4x\Rightarrow x>\frac 12$

Thus the values for $x$, such that$ \ log_{(1-|x|)}|(3x-1)|<1$ are $\frac 12<x<1$

  • $\begingroup$ Apologies for the original solution, I think I am going crazy haha. Cheers! $\endgroup$ Apr 8 '17 at 6:03
  • $\begingroup$ Wow! Thanks a lot. You really helpful. Could you tell me which part i got wrong in my work? $\endgroup$ Apr 8 '17 at 7:01
  • $\begingroup$ You said the opposite, that is $(1-|x|)>(3x-1)$. Instead, first inverse the original logarithm into its exponential equivalent. (Take the logarithm's base to the power of the RHS (1), and set it equal to $3x-1$). Don't forget to accept and up vote this answer. Hope this help! $\endgroup$ Apr 8 '17 at 7:12
  • $\begingroup$ Done! Thanks for the answer. $\endgroup$ Apr 8 '17 at 7:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.