Simple inequality with natural logarithm












-1














I need to calculate such inequality.



$ lnfrac{2x}{x-1} geq 0 $



I'm new to concept of $ ln$ and clueless how to move on. Any tips?






logarithms







share|cite|improve this question
























  • Do you at least know for which $x$ the formula is well defined?
    – Federico
    Nov 21 '18 at 19:40










  • Hint: this is equivalent to $frac{2x}{x-1}ge1$.
    – Yves Daoust
    Nov 21 '18 at 19:41
















-1














I need to calculate such inequality.



$ lnfrac{2x}{x-1} geq 0 $



I'm new to concept of $ ln$ and clueless how to move on. Any tips?






logarithms







share|cite|improve this question
























  • Do you at least know for which $x$ the formula is well defined?
    – Federico
    Nov 21 '18 at 19:40










  • Hint: this is equivalent to $frac{2x}{x-1}ge1$.
    – Yves Daoust
    Nov 21 '18 at 19:41














-1












-1








-1







I need to calculate such inequality.



$ lnfrac{2x}{x-1} geq 0 $



I'm new to concept of $ ln$ and clueless how to move on. Any tips?






logarithms







share|cite|improve this question















I need to calculate such inequality.



$ lnfrac{2x}{x-1} geq 0 $



I'm new to concept of $ ln$ and clueless how to move on. Any tips?






logarithms




logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 '18 at 19:45









Bernard

118k639112




118k639112










asked Nov 21 '18 at 19:37









wenoweno

1078




1078












  • Do you at least know for which $x$ the formula is well defined?
    – Federico
    Nov 21 '18 at 19:40










  • Hint: this is equivalent to $frac{2x}{x-1}ge1$.
    – Yves Daoust
    Nov 21 '18 at 19:41


















  • Do you at least know for which $x$ the formula is well defined?
    – Federico
    Nov 21 '18 at 19:40










  • Hint: this is equivalent to $frac{2x}{x-1}ge1$.
    – Yves Daoust
    Nov 21 '18 at 19:41
















Do you at least know for which $x$ the formula is well defined?
– Federico
Nov 21 '18 at 19:40




Do you at least know for which $x$ the formula is well defined?
– Federico
Nov 21 '18 at 19:40












Hint: this is equivalent to $frac{2x}{x-1}ge1$.
– Yves Daoust
Nov 21 '18 at 19:41




Hint: this is equivalent to $frac{2x}{x-1}ge1$.
– Yves Daoust
Nov 21 '18 at 19:41










1 Answer
1






active

oldest

votes


















-1














Hint:




  1. You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
    $$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$

  2. In this domain of validity, the log is non-negative if and only if
    $$frac{2x}{x-1}ge 1. $$






share|cite|improve this answer





















  • Thanks. May I ask where the 2x/(x-1) >= 1 came from?
    – weno
    Nov 21 '18 at 20:11










  • $ln 1=0$, and the logarithm is an increasing function.
    – Bernard
    Nov 21 '18 at 20:14










  • The maniac downvoter struck again!
    – Bernard
    Nov 22 '18 at 1:12











Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1














Hint:




  1. You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
    $$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$

  2. In this domain of validity, the log is non-negative if and only if
    $$frac{2x}{x-1}ge 1. $$






share|cite|improve this answer





















  • Thanks. May I ask where the 2x/(x-1) >= 1 came from?
    – weno
    Nov 21 '18 at 20:11










  • $ln 1=0$, and the logarithm is an increasing function.
    – Bernard
    Nov 21 '18 at 20:14










  • The maniac downvoter struck again!
    – Bernard
    Nov 22 '18 at 1:12
















-1














Hint:




  1. You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
    $$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$

  2. In this domain of validity, the log is non-negative if and only if
    $$frac{2x}{x-1}ge 1. $$






share|cite|improve this answer





















  • Thanks. May I ask where the 2x/(x-1) >= 1 came from?
    – weno
    Nov 21 '18 at 20:11










  • $ln 1=0$, and the logarithm is an increasing function.
    – Bernard
    Nov 21 '18 at 20:14










  • The maniac downvoter struck again!
    – Bernard
    Nov 22 '18 at 1:12














-1












-1








-1






Hint:




  1. You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
    $$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$

  2. In this domain of validity, the log is non-negative if and only if
    $$frac{2x}{x-1}ge 1. $$






share|cite|improve this answer












Hint:




  1. You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
    $$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$

  2. In this domain of validity, the log is non-negative if and only if
    $$frac{2x}{x-1}ge 1. $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 '18 at 19:54









BernardBernard

118k639112




118k639112












  • Thanks. May I ask where the 2x/(x-1) >= 1 came from?
    – weno
    Nov 21 '18 at 20:11










  • $ln 1=0$, and the logarithm is an increasing function.
    – Bernard
    Nov 21 '18 at 20:14










  • The maniac downvoter struck again!
    – Bernard
    Nov 22 '18 at 1:12


















  • Thanks. May I ask where the 2x/(x-1) >= 1 came from?
    – weno
    Nov 21 '18 at 20:11










  • $ln 1=0$, and the logarithm is an increasing function.
    – Bernard
    Nov 21 '18 at 20:14










  • The maniac downvoter struck again!
    – Bernard
    Nov 22 '18 at 1:12
















Thanks. May I ask where the 2x/(x-1) >= 1 came from?
– weno
Nov 21 '18 at 20:11




Thanks. May I ask where the 2x/(x-1) >= 1 came from?
– weno
Nov 21 '18 at 20:11












$ln 1=0$, and the logarithm is an increasing function.
– Bernard
Nov 21 '18 at 20:14




$ln 1=0$, and the logarithm is an increasing function.
– Bernard
Nov 21 '18 at 20:14












The maniac downvoter struck again!
– Bernard
Nov 22 '18 at 1:12




The maniac downvoter struck again!
– Bernard
Nov 22 '18 at 1:12


















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