Logarithmic Inequations

[timo]

Ok, I think I understand your problem now. The inequality is not

[math]\ln(\frac{x-1}{x+2}) > 0[/math]

but

[math]\ln (x-1) - \ln (x+2) > 0 [/math].

The rearranged inequality has a larger definition range than the original inequality. We hopefully agree that the original inequality is not defined for x<1. As a related example: f(x)=x/x is not the same function as g(x)=1.

[NeonBlack]

Urm....

 

I didn't make up this example and take another look at it and you'll see that IT IS bigger than 1 (at least according to my book!)

 

Let's just assume NeonBlack was posting after too little sleep or something...

 

Did I miss something? I was definitely up too late last night, so it's possible.

[thedarkshade]

Ok, I think I understand your problem now. The inequality is not

[math]\ln(\frac{x-1}{x+2}) > 0[/math]

but

[math]\ln (x-1) - \ln (x+2) > 0 [/math].

It is the same thing!

 

Ok, I think I understand your problem now. The inequality is not

[math]\ln(\frac{x-1}{x+2}) > 0[/math]

but

[math]\ln (x-1) - \ln (x+2) > 0 [/math].

The rearranged inequality has a larger definition range than the original inequality. We hopefully agree that the original inequality is not defined for x<1. As a related example: f(x)=x/x is not the same function as g(x)=1.

In a more simple way I think me must put [math] -\ln (x+2)[/math] on the other side so it turns positive and then it goes like this:

[math]\ln x-1 > \ln x+2[/math] and then we drop the logs and as the base is bigger than 1 the inequality sign doesn't change so we get three inequations

[math]x-1 > x+2[/math]

[math]x-1 > 0[/math]

[math]x+2 > 0[/math] the last two equations are because log of negative numbers does not exist!

And then all we got to do is find the values, place them on the number line and the interval that matches with the sign of the inequality is (or are) the solution!

[Country Boy]

I watched some videos lately which were really helpful and according to them the equation I posted earlier has to be solved this way:

 

logx-1 - logx+2 > 0

 

logx-1 > logx+2 (drop the logs) and we get

 

x-1 > x+2 and x-1>0 and also x+2>0

 

for x-1>0 get clearly get x>1

for x+2>0 we get x>-2

 

but for x - 1 > x + 2 (x on both sides so they're canceled)

we get -1>2 ; which is impossible so this falls ( at least I think so!)

 

next what we got to do is draw the number line and but x>1 and x>-2 there, and as the sign is greater (>) then we take positive values.

 

I think it must be done this way.

what do you guys think?

No, I don't think it MUST be done that way. In fact, the solution I THINK you are giving, x> 1, is clearly incorrect. There are many different ways to do a problem like that. What I would have done was immediately combine the logarithms:

log(x-1)- log(x+2)>0 is the same as log((x-1)/(x+2))> 0 and that means, of course that (x-1)/(x+2)> 1. Now use the advice you were given originally: think it this as an equation: (x-1)/(x+2)= 1 is the same as x-1= x+ 2 which has no solution. The point of "treat it as an equation" is that the only place a CONTINUOUS function can change from "< 1" to ">1" is where it is EQUAL to 1. Since x-1= x+2 is impossible that never happens. However, the function (x-1)/(x+2) is NOT continuous at x= -2. Of course, the orginal function log(x-1) is only defined for x> 1. If x= 2, say, then log(x-1)-log(x+2) becomes log(1)- log(3) which is negative. log(x-1)- log(x+2)< 0 for all x> 1 and is undefined for any other x.

 

The original inequality, log(x-1)- log(x+2)> 0 is NEVER true.

[the tree]

While I can't really see anything wrong in what HallsOfIvy says, Maple seems to disagree, I'll put that down to a fault on Maple's part I guess.

[thedarkshade]

The original inequality, log(x-1)- log(x+2)> 0 is NEVER true.

What do you mean NEVER true???!!!

Like there can't be a number (for x) so that the inequality would be valid?

[the tree]

I'm fairly sure that's what he's saying.

[thedarkshade]

I'm fairly sure that's what he's saying.

Well that's wrong then, cause it does have a solution.

You can see that just looking at the inequation, without doing it at all!!

[timo]

Perhaps you should enlighten us and tell us a solution.

[thedarkshade]

Perhaps you should enlighten us and tell us a solution.

 

sure, -infinite<x<-2

[timo]

log(-5) = ?

[thedarkshade]

log(-5) = ?

I asked myself too about that, and I agree, you are completely right, but I got this PhD book that says what I previously posted.

[timo]

That is because your PhD book might define [math] \log (-5) = \log(5) + i\pi [/math], in which case then [math]\log(-5) - \log(-2) = \log(5) + i\pi - \left( \log(2) + i\pi \right) = \log(5) - \log(2) [/math]. This does indeed look like being >0. But strictly speaking you have complex numbers on the left-hand side of the inequality (even though their imaginary parts will often cancel). The relations ">" and "<" are not (canonically) defined for complex numbers. Advanced-level books often make inherit assumptions that are not explicitely mentioned; without knowing the book or the context (no need to quote, no one on sfn, safe for possibly two or three people, will understand math on a PhD level).

[thedarkshade]

no one on sfn, safe for possibly two or three people, will understand math on a PhD level.

I guess I'm one of you too. I spend pages and pages and pages on that drill, and yet it always turned the same result:doh: .

Thnx

[the tree]

Whist I also got [math]x \in (-\infty , -2)[/math], (did I say that? I forget.) I can see no flaw in HallsOfIvy's reasoning and unless you can find the flaw that trumps any appeal non specific authority.

[thedarkshade]

Whist I also got [math]x \in (-\infty , -2)[/math], (did I say that? I forget.) I can see no flaw in HallsOfIvy's reasoning and unless you can find the flaw that trumps any appeal non specific authority.

 

So you can that's it? That's the solution?

[the tree]

No, I think HallsOfIvy was right (and thus that I was wrong), I think it has no solutions.

[thedarkshade]

Wait a minute,, of course [math]

x \in (-\infty , -2)

[/math] is the solution. I mean take -3 for example and we'd get:

[math]

\ln(\frac{-3-1}{-3+2}) > 0

[/math]

and then:

[math]

\ln(\frac{-4}{-1}) > 0

[/math]

minus on both sides of fraction gives you plus, so we get:

[math]

\ln(4) > 0

[/math] and that's it!

[the tree]

I said that a while back, and HallsOfIvy explained why it was wrong. There are no solutions.

[thedarkshade]

But HallsOfIvy was referring to the equation [math]\ln{(x-1)} - \ln{(x+2)} > 0[/MATH] but the real equation is:[math]

\ln(\frac{x-1}{x+2}) > 0

[/math] and my teacher said it's right, and IT IS right! he's got a PhD you know!!!

[ydoaPs]

[math]\ln(x-1)-\ln(x+2)=\ln(\frac{x-1}{x+2})[/math]

[thedarkshade]

[math]\ln(x-1)-\ln(x+2)=\ln(\frac{x-1}{x+2})[/math]

 

yes ,that's right but the solution [math]

x \in (-\infty , -2)

[/math] doesn't work for the left side, because you get log of negative numbers, but it does work for the right side!

[the tree]

But HallsOfIvy was referring to the equation [math]\ln{(x-1)} - \ln{(x+2)} > 0[/MATH] but the real equation is:[math]

\ln(\frac{x-1}{x+2}) > 0

[/math]

We were talking about the one that we were originally presented with. You can't say we're wrong because you changed the question retroactively.

and my teacher said it's right, and IT IS right! he's got a PhD you know!!!

Appeals to authority have no meaning in mathematics. If Newton jumped out of his grave and told me that it was right, I'd still expect a reason from him beyond "IT IS RIGHT DAMNIT".

[Country Boy]

But HallsOfIvy was referring to the equation [math]\ln{(x-1)} - \ln{(x+2)} > 0[/MATH] but the real equation is:[math]

\ln(\frac{x-1}{x+2}) > 0

[/math] and my teacher said it's right, and IT IS right! he's got a PhD you know!!!

 

The "real" equation? What you originally said was

Thanks for all that guys, but could me get to something more practical.

I'll take a very very simple example and you just tell how ti goes:

 

logx-1 -logx+2>0

then we get logx-1/x+2>0

and according to the rules we learned x-1/x+2>0

Now which equation did you ask your teacher about? The real inequality you originally gave us was log(x-1)- log(x+2)> 0

 

The inequality log((x-1)/(x+2))> 0 is satisfied as long as (x-1)/(x+2)> 1.

That is the same as (x-1)/(x+2)-(x+2)/(x+2)= -3/(x+2)> 0, which, in turn, is true as long as x+2 is negative: x+ 2< 0 or x< -2.

 

However, the original inequality, as you give above, was log(x-1)- log(x+2)> 0. It can only have, as solutions, the solutions to the "equivalent" log((x-1)/(x+2))> 0 and those only if they satisfy the original inequality. Since any x< -2 makes the arguments of both logarithms negative, there is NO solution to your equation.

 

I am sure your teacher is a brilliant man. Unfortunately, he can only answer the question you ask, which may not be the correct one!

[thedarkshade]

Listen HallsOfIvy.

there are different ways of solving logarithmic inequations, and what I did above what just applying one rule [math]log\frac{x}{y}=logx - logy[/math] to that equation!

if the solution [math]

x \in (-\infty , -2)

[/math] doesn't work for [math]log(x-1) - log(x+2)>0[/math] then just switch the equation to [math]log\frac{x-1}{x+2}[/math]