How to solve this logarithm inequality with absolute value as its base?

Question

How to deal with this ?

$$\log_{|1 - x|} (x+5)>2 $$

the $|1-x|$ is the base of the logarithm.

I tried this below approach but it seems not the complete solution. \begin{align} \frac{\log(x+5)}{\log|1-x|} & > 2\\ {\log(x+5)} & > 2{\log|1-x|}\\ {\log(x+5)} & > {\log|1-x|^2}\\ (x+5) & >|1-x|^2\\ (x+5) & >(1-x)^2\\ (x+5) & >1-2x+x^2\\ x^2-3x-4& < 0 \end{align} $$ -1<x<4 $$ I also checked with wolframalpha. https://www.wolframalpha.com/input/?i=log+%5Babsolut(1-x),+(x%2B5)%5D%3C2

I appreciate your help.

You can't just multiply the inequality by $\log|1-x|$ you must first divide it into cases when it's positive and when it's negative — kingW3, Mar 18 '17 at 20:02

Sid · Accepted Answer · 2017-03-25T18:37:04.693

2

Look at the pictures. You will see there are two cases to the problem.

Here is a link to : Wolfram Alpha

Hope this helps. If you need help in solving the individual inequalities, comment and I will show this as well.

edited Mar 25 '17 at 18:37

answered Mar 25 '17 at 18:26

Sid

1,234

Please type out your equations. – GNUSupporter 8964民主女神地下教會 Mar 28 '17 at 07:38
sorry. I'm new so still getting familiar with Latex – Sid Mar 28 '17 at 07:40
For some basic information about writing math at this site see e.g. here, here, here and here. Btw, it's not "graph theory". – GNUSupporter 8964民主女神地下教會 Mar 28 '17 at 07:44
yeah sorry. graph theory is another thing i realized someone told me. – Sid Mar 28 '17 at 07:48
Very nice solution +1 – Navin Mar 30 '17 at 09:29
@navinstudent thank you. I saw you saw my similar but more complicated question. Thanks in advance for providing a general solution to my question. And thanks for editing my question to make it more clear. – Sid Mar 30 '17 at 09:56
I didnot edit your question.It was @Did who edited the question.I also find your question very complicated. – Navin Mar 30 '17 at 09:58
@Sid, My little brother and I thank you for helping us. Now, I can see the problem much more clearly. – Codelearner777 Apr 07 '17 at 17:43
@Sid, My little brother and I thank you for helping us. Now, I can see the problem much more clearly.
one question.. about $$ 0<(x+5)/(1-x)^2 <1 $$

on this part $ 0<(x+5)/(1-x)^2 $, isn't the solution should be $ x>-5, x<1, x>1 $?

I also checked with wolframalpha.

http://www.wolframalpha.com/input/?i=Solve+the+inequality:+%5B%2F%2Fmath:0+%3C(x%2B5)%2F(1-x)%5E2%2F%2F%5D

could you guide me one more time?
– Codelearner777 Apr 07 '17 at 17:50
@codelearner777 you are correct. This is a standard method to solve logarithmic inequalities with a variable base. You need to check two regions. – Sid Apr 08 '17 at 04:29
i recommend you to solve this: http://math.stackexchange.com/questions/1422858/solve-logarithmic-equation-log-fracx5x2-8x16-geq-0 when you are finished sucessfully comment here. and I will give on more problem to clear ALL concepts – Sid Apr 08 '17 at 12:01

score 0 · Answer 2 · answered Mar 18 '17 at 20:02

0

Hint: Note that the base should be $$\left| 1-x \right| \neq 1\Rightarrow x\neq 0\quad \& \quad x\neq 2$$ and $$x+5>0\quad \Rightarrow \quad x>-5$$

answered Mar 18 '17 at 20:02

haqnatural

21,578

finding the domain of the expression is not helpful. – Sid Apr 07 '17 at 19:15

GNUSupporter 8964民主女神地下教會 · Answer 3 · 2017-03-18T20:27:39.407

0

Developping kingW3's idea, we'll know why that's empty region in the solution. We first note that $x \ne 1$.

The problem in the question body is the transition from $$\frac{\log(x+5)}{\log|1-x|} > 2\tag1\label1$$ to $${\log(x+5)} > 2{\log|1-x|}.\tag2\label2$$

By neglecting case 2 below, we failed to capture the empty region $0 < x < 2$ in the range for $x$.

Case 1: $\log|1-x|>0 \iff |1-x|>1 \iff x<0$ or $x>2$. Using OP's calculations above, we get $-1 < x < 0$ or $2 < x < 4$, which agrees with the graph of the linked Wolfram Alpha page.

Case 2: $\log|1-x|<0 \iff 0<|1-x|<1 \iff 0 < x < 1$ or $1 < x < 2$.

We should reverse the inequality \eqref{2} in this case since we're multiplying \eqref{1} by a negative denominator $\log|1-x|$.

$${\log(x+5)} < 2{\log|1-x|}$$

As a result, the inequalities in the question body that follow from \eqref{2} should be reversed, until $x^2-3x-4>0$, from which we deduce $x<-1$ or $x>4$. The intersection of $(-\infty,-1)\cup(4,\infty)$ with $(0,1)\cup(1,2)$ is empty, so no real value of $x$ satisfies case 2.

Hence the solution is $-1<x<0$ or $2<x<4$.

edited Mar 18 '17 at 20:27

answered Mar 18 '17 at 20:22

GNUSupporter 8964民主女神地下教會

17,627

this is the right method but it is not working with my problem. See here: [http://math.stackexchange.com/questions/2201633/how-to-solve-this-logarithmic-inequality-with-another-logarithm-as-its-base][1] – Sid Mar 28 '17 at 05:57
@Sid Thx for referring me to your new problem. I hope someone else can solve it for you. – GNUSupporter 8964民主女神地下教會 Mar 28 '17 at 07:47
I dont want a complete solution. I can solve all the inequalities myself. I just need a general solution. – Sid Mar 28 '17 at 07:56
@Sid Though I don't see one, I hope someone else can give it to you. – GNUSupporter 8964民主女神地下教會 Mar 28 '17 at 08:59
I'll wait. Maybe you can try I again. I edited my question and gave it more context. I explained a similar inequality and wrote the general solution. I just need one for the more complicated one now. I hope so too, I'll wait for someone to give it. – Sid Mar 28 '17 at 09:12

How to solve this logarithm inequality with absolute value as its base?

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