Mixing
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
edited Nov 21 '18 at 19:45
Bernard
118k639112
asked Nov 21 '18 at 19:37
wenoweno
1078
add a comment |
-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
edited Nov 21 '18 at 19:45
Bernard
118k639112
asked Nov 21 '18 at 19:37
wenoweno
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
add a comment |
-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
edited Nov 21 '18 at 19:45
Bernard
118k639112
asked Nov 21 '18 at 19:37
wenoweno
1078
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
edited Nov 21 '18 at 19:45
Bernard
118k639112
asked Nov 21 '18 at 19:37
wenoweno
1078
edited Nov 21 '18 at 19:45
Bernard
118k639112
asked Nov 21 '18 at 19:37
wenoweno
1078
edited Nov 21 '18 at 19:45
Bernard
118k639112
edited Nov 21 '18 at 19:45
Bernard
118k639112
edited Nov 21 '18 at 19:45
Bernard
118k639112
118k639112
asked Nov 21 '18 at 19:37
wenoweno
1078
asked Nov 21 '18 at 19:37
wenoweno
1078
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
-1
Hint:
- You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
$$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$
- In this domain of validity, the log is non-negative if and only if
$$frac{2x}{x-1}ge 1. $$
answered Nov 21 '18 at 19:54
BernardBernard
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
-1
Hint:
- You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
$$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$
- In this domain of validity, the log is non-negative if and only if
$$frac{2x}{x-1}ge 1. $$
answered Nov 21 '18 at 19:54
BernardBernard
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
add a comment |
-1
Hint:
- You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
$$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$
- In this domain of validity, the log is non-negative if and only if
$$frac{2x}{x-1}ge 1. $$
answered Nov 21 '18 at 19:54
BernardBernard
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
add a comment |
-1
-1
-1
Hint:
- You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
$$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$
- In this domain of validity, the log is non-negative if and only if
$$frac{2x}{x-1}ge 1. $$
answered Nov 21 '18 at 19:54
BernardBernard
118k639112
Hint:
- You have to specify the domain of validity of this inequation: $dfrac{2x}{x-1}$ has to be defined and $>0$. This means
$$frac{x}{x-1};text{ defined and }>0iff x(x-1)>0. $$
- In this domain of validity, the log is non-negative if and only if
$$frac{2x}{x-1}ge 1. $$
answered Nov 21 '18 at 19:54
BernardBernard
118k639112
answered Nov 21 '18 at 19:54
BernardBernard
118k639112
answered Nov 21 '18 at 19:54
BernardBernard
118k639112
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
add a comment |
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
add a comment |
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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