Solve inequality logarithm

Question

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.

Answer 1

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 $\frac 12<x<1$