Log _{x} ((x+3)/(x-1)) > Log _{x} x ?? how to find x that satisfy this inequality?
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Originally Posted by psw1412 Log _{x} ((x+3)/(x-1)) > Log _{x} x ?? how to find x that satisfy this inequality? Solve the inequality and choose the interval in which x > 0
Originally Posted by votan Solve the inequality and choose the interval in which x > 0 is it correct??
Last edited by psw1412; Sep 22nd 2013 at 11:12 AM.
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 $\displaystyle x<-1\text{ or }1<x<3$
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: 2x=4 x=2
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
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.
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.
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