1. ## Logharitmic inequality

$\displaystyle log_x|y-2|>log_{|y-2|} x$, Can somebody help me?

2. ## Re: Logharitmic inequality

Originally Posted by greg1987
$\displaystyle log_x|y-2|>log_{|y-2|} x$, Can somebody help me?
It looks like you have dumped us into the middle of a problem.
What is the exact statement of the entire problem?
There must be more to the question.

3. ## Re: Logharitmic inequality

hi

try using: $\displaystyle \log_a b=\frac{1}{\log_b a}$

then use substitution.

hope this helps.

4. ## Re: Logharitmic inequality

We've got to find a set of points (x,y) (and mark it on a plane) which satisfies this inequality.That's all.

5. ## Re: Logharitmic inequality

Originally Posted by anonimnystefy
hi

try using: $\displaystyle \log_a b=\frac{1}{\log_b a}$

then use substitution.

hope this helps.
ok so i've got $\displaystyle \log_x |y-2|>\frac{1}{\log_x |y-2|}$
Now I need to multiply the denominator?

6. ## Re: Logharitmic inequality

you cannot multiply because you don't know if the sign of the denominator is + or -.

did you try using substitution?it will make it a little easier to see what you can do.

7. ## Re: Logharitmic inequality

I don't know what should I substitute, some clue?

8. ## Re: Logharitmic inequality

try $\displaystyle t=\log_x |y-2|$

9. ## Re: Logharitmic inequality

so -1<t<0 or t>1?

10. ## Re: Logharitmic inequality

that's pretty much right.

what do you get when you sub?

11. ## Re: Logharitmic inequality

rewrite $log_x|y-2|>log_{|y-2|} x$ by meaning:

x^a = (y-2); (y -2)^(1/a) = x

$log_x|y-2|>log_{|y-2|} x$ is same as:

a > 1/a solve for a

12. ## Re: Logharitmic inequality

$\displaystyle t=\log_x |y-2|>1$ or $\displaystyle -1<t=\log_x |y-2|<0$ and how can I determine x and y?

13. ## Re: Logharitmic inequality

Originally Posted by greg1987
We've got to find a set of points (x,y) (and mark it on a plane) which satisfies this inequality.That's all.
Well then you have included that.
You know that $\displaystyle x>0~\&~x\ne 1$, WHY?
For the same reason $\displaystyle y\ne 2$.
If $\displaystyle y>e+2$ then $\displaystyle \log_x|y-2|>0$ WHY?
What about $\displaystyle y<2-e~?$

There is a start.

14. ## Re: Logharitmic inequality

Here is a plot (starting from 100,000 random points), both axes are [-5,5].