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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.

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  • 1
    $\begingroup$ 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 $\endgroup$ – kingW3 Mar 18 '17 at 20:02
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enter image description here

enter image description here

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.

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

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  • $\begingroup$ finding the domain of the expression is not helpful. $\endgroup$ – Sid Apr 7 '17 at 19:15
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Developping kingW3's idea, we'll know why that's empty region in the solution. We first note that $x \ne 1$.

wolfram alpha graph

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 $2<x<4$.

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  • $\begingroup$ this is the right method but it is not working with my problem. See here: [math.stackexchange.com/questions/2201633/… $\endgroup$ – Sid Mar 28 '17 at 5:57
  • $\begingroup$ @Sid Thx for referring me to your new problem. I hope someone else can solve it for you. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 28 '17 at 7:47
  • $\begingroup$ I dont want a complete solution. I can solve all the inequalities myself. I just need a general solution. $\endgroup$ – Sid Mar 28 '17 at 7:56
  • $\begingroup$ @Sid Though I don't see one, I hope someone else can give it to you. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 28 '17 at 8:59
  • $\begingroup$ 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. $\endgroup$ – Sid Mar 28 '17 at 9:12

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