Hi everyone!

 

Lately at school we've been dealing with logarithmic inequations and they seem pretty tricky. So I was wondering if anyone could add a link or just post some useful hints that would help solving these inequations.

 

I'd appreciate any kind of help!

Pretend they're equations until the end, iirc.

Pretend they're equations until the end, iirc.

 

How possible is that! You can't multiply with a variable like you do with equations!

You can't? What do you mean "multiply with a variable"? The only real differences I remember are how to graph them (solid line v dotted and where to shade) and multiplying or dividing by a negative(flip the inequality from < to > or vice versa).

He means with something like g(x)>f(x) you can't multiply by x, since sometimes it has negative values and sometimes it doesn't.

 

The pretend they're equations technique still wins, you just have a smaller arsenal of operations to work with.

 

Sketching a graph often helps, sometimes just so you know how many points your looking for and to get some intuition about which way round to put the inequality.

He means with something like g(x)>f(x) you can't multiply by x, since sometimes it has negative values and sometimes it doesn't.

 

Couldn't you just have three solutions(one where x<0, one were x=0, and one where x>0)?

Then you're doing three problems instead of one, why would you inflict that upon yourself?

Then you're doing three problems instead of one, why would you inflict that upon yourself?

But would it work?

It wouldn't always be those constraints (for instance if you were multiplying through by (x³-3) you'd want entirely different ones) but in principle yes it would.

Sweet! I am super-awesome!

 

How would you go about it algebraically without having different inequalities for different sections of the analyzed range?

 

 

Wait....couldn't you just set f(x)=g(x) to find the x values where they intersect? Then plug in x values between intersection x values into both equation to see which equation is greater than the other for which range?

 

thedarkshade, if you could provide a sample problem, it would probably help us help you as we'd know more precisely what you are asking.

Assuming that both sides represent continuous functions that exist in the same range, yeah, that's basically what you were saying at the beginning of the thread and you weren't wrong. Otherwise you just need to do that and then knock off values of x where there's no value of f(x), which is easy.

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

 

and that were it come what I said that you can't multiply with a variable (in this case (x-2).

So what should I do here? Equalize the left side with a zero????

 

Or it could be this way

 

logx-1 -logx+2>0

logx-1>logx+2

and then we'd get x-1>0

x+2>0

x-1>x+2

 

Please Help me!!!

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

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

This is not quite right.

I am assuming you know that [math]e^{\ln{u}}=u[/math]

this means that you want [math]\frac{x-1}{x+2}>1[/math] (since [math]e^0=1[/math])

Maybe this was just an example you made up' date=' but [math']\frac{x-1}{x+2}[/math] cannot be larger than 1.

 

Edit:

I should clarify that this is true for a log of any base b. [math]b^{\log_b{u}}=u[/math]

Maybe this was just an example you made up, but [math]\frac{x-1}{x+2}[/math] cannot be larger than 1.

Urm....

Maybe this was just an example you made up, but [math]\frac{x-1}{x+2}[/math] cannot be larger than 1.

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, everyone makes mistakes like that from time to time.

 

In this case, here's what you want to do:

1. Consider what conditions are necessary for [math]\ln(y)[/math] to have a value at all.
   
2. Consider what conditions are necessary for [math]\ln(y)[/math] to have a value greater than one.
   
3. So, you've got some bounds on [math]y[/math], now let [math]y=\frac{x-1}{x+2}[/math], and solve your new inequality the old fashioned way (like YDOAP said).

Let's just assume NeonBlack was posting after too little sleep or something, everyone makes mistakes like that from time to time.

Actually, I wonder how you want to get any solution without violating the range, too. The original inequality obviously is defined for x>1, only. I spontaneously see no x>1 for which (x-1)/(x+2) > 1.

The original inequality obviously is defined for x>1, only.

What do you mean by obviously? The original inequality was

[math]\ln(x-1) - \ln(x+2) >0[/math], no mention was made of range.

 

edit: I'm aware the log in this case isn't defined for [math]x \in (-2,1)[/math] but none of that interval was proposed anyway.

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?

I think it must be done this way.

what do you guys think?

I think that you should pick 3 different values of x (i.e. -3, -1, 2) and see if they fit with your answer.

What do you mean by obviously? The original inequality was

[math]\ln(x-1) - \ln(x+2) >0[/math], no mention was made of range.

 

edit: I'm aware the log in this case isn't defined for [math]x \in (-2,1)[/math] but none of that interval was proposed anyway.

The inequality isn't defined for x<-2, either. The range probably wasn't mentioned because it is "obvious" .

 

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

That is correct. Do you know why you can drop the logs ?

 

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!)

Yes.

 

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 don't understand what you are saying here.

 

I think it must be done this way.

what do you guys think?

You can also do it the x-1/x+2>1 as long as you remember for which x the original equation was defined. It would also show that no x>1 satisfies the inequality.

What I meant Atheist about drawing number line I meant to draw OX line and put the number we found in there and them take positive side as solutions as the sign was >.

The inequality isn't defined for x<-2, either. The range probably wasn't mentioned because it is "obvious" .

Could you please actually explain that? There are without a doubt values for that log, so why not for the whole inequality?

Can you demonstrate that it is not defined?

Could you please actually explain that?

Sure.

There are without a doubt values for that log, ...

I have a lot of doubts that the logarithm of a negative number is defined.

...so why not for the whole inequality?

Because if you try to be smart and use a definition of log (which one would you use considering that none is given?) that maps negative numbers on complex values, then the ">" sign is not defined anymore.

Can you demonstrate that it is not defined?

Put "log(-1)" in the calculator of your choice.

I have a lot of doubts that the logarithm of a negative number is defined.

 

Put "log(-1)" in the calculator of your choice.

I still don't understand. What has the logarithm of a negative number got to do having a negative value of x in [math]\ln(\frac{x-1}{x+2})[/math]? If say x were -3, that'd be [math]\ln(4)[/math], would you claim that [math]\ln(4)[/math] is undefined?