# Logarithm Inequality. Help!

• Sep 22nd 2013, 09:13 AM
psw1412
Logarithm Inequality. Help!
Log x ((x+3)/(x-1)) > Log x x ??

how to find x that satisfy this inequality?
• Sep 22nd 2013, 09:39 AM
votan
Re: Logarithm Inequality. Help!
Quote:

Originally Posted by psw1412
Log x ((x+3)/(x-1)) > Log x x ??

how to find x that satisfy this inequality?

Attachment 29243

Solve the inequality and choose the interval in which x > 0
• Sep 22nd 2013, 11:03 AM
psw1412
Re: Logarithm Inequality. Help!
Quote:

Originally Posted by votan
Attachment 29243

Solve the inequality and choose the interval in which x > 0

is it correct??
• Sep 22nd 2013, 11:23 AM
Plato
Re: Logarithm Inequality. Help!
Quote:

Originally Posted by psw1412
is it correct??

NO! It should be:
$\displaystyle \\\frac{3-x}{x-1}>x\\\frac{3-x}{x-1}-x>0\\\frac{3+2x-x^2}{x-1}>0\\\frac{(3-x)(1+x)}{x-1}>0$.

That means $\displaystyle x<-1\text{ or }1<x<3$
• Sep 22nd 2013, 11:57 AM
psw1412
Re: Logarithm Inequality. Help!
thank you.
one more question. How could you write that inequality from codecogs.com in a few rows? Every time I press "enter" it continues in the same row. Like this: Attachment 29245

2x=4
x=2
• Sep 22nd 2013, 01:56 PM
votan
Re: Logarithm Inequality. Help!
Quote:

Originally Posted by Plato
NO! It should be:
$\displaystyle \\\frac{3-x}{x-1}>x\\\frac{3-x}{x-1}-x>0\\\frac{3+2x-x^2}{x-1}>0\\\frac{(3-x)(1+x)}{x-1}>0$.

That means $\displaystyle x<-1\text{ or }1<x<3$

The argument of log cannot be negative. Also, the numerator is x + 3, not 3 - x
• Sep 22nd 2013, 01:58 PM
Plato
Re: Logarithm Inequality. Help!
Quote:

Originally Posted by votan
The argument of log cannot be negative. Also, the numerator is x + 3, not 3 - x

I was not addressing the OP.
If I were then I would have quoted it.
• Sep 22nd 2013, 02:01 PM
votan
Re: Logarithm Inequality. Help!
Quote:

Originally Posted by psw1412
is it correct??

Do not multiply both sides by the denominator. It is OK in equality provided the value of x does not make the denominator 0.