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.
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Please have a look at the introduction to posting mathematical expressions. You will find there links to more advanced topics. – hardmath Apr 08 '17 at 06:02
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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. – it's a hire car baby Apr 08 '17 at 07:14
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Sorry. I can't put mathematic symbol here. i don't know how. – Fauzy Nur Noviansyah Apr 08 '17 at 07:30
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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 $\frac 12<x<1$
Mark Pineau
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Apologies for the original solution, I think I am going crazy haha. Cheers! – Mark Pineau Apr 08 '17 at 06:03
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Wow! Thanks a lot. You really helpful. Could you tell me which part i got wrong in my work? – Fauzy Nur Noviansyah Apr 08 '17 at 07:01
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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! – Mark Pineau Apr 08 '17 at 07:12
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