Mixing
Why are $ln x$ and $e^x$ considered to be each others' inverses?
From my understanding the definition of a function's inverse is as follows.
Take a function $f$ which has the inverse $f^{-1}$. This would mean that $f(f^{-1}(x)) = x$ and that $f^{-1}(f(x)) = x$ for every real value for $x$. Right? And this is true for $ln(e^x) = x$ but for $e^{(ln x)} = x$ it only holds true for $x > 0$. Why are they still considered each others' inverse?
real-analysis inverse-function
edited Nov 21 '18 at 17:48
Rócherz
2,7762721
asked Nov 21 '18 at 17:41
CrazyMineCoderCrazyMineCoder
32
add a comment |
From my understanding the definition of a function's inverse is as follows.
Take a function $f$ which has the inverse $f^{-1}$. This would mean that $f(f^{-1}(x)) = x$ and that $f^{-1}(f(x)) = x$ for every real value for $x$. Right? And this is true for $ln(e^x) = x$ but for $e^{(ln x)} = x$ it only holds true for $x > 0$. Why are they still considered each others' inverse?
real-analysis inverse-function
edited Nov 21 '18 at 17:48
Rócherz
2,7762721
asked Nov 21 '18 at 17:41
CrazyMineCoderCrazyMineCoder
32
1
It's true because the range of $xmapsto e^x$ for $xin mathbb{R}$ is $(0, infty)$, so of course it only makes sense for the inverse to be defined on $(0, infty)$. Strictly speaking, your definition of inverse isn't quite right; the inverse of a function $f : Xto Y$ is defined $f^{-1} : Yto X$.
– Michael Lee
Nov 21 '18 at 17:46
As @Micheal Lee explained, if f:A→B then it's inverse is f−1:B→A, and these sets are not necessarily the all the real numbers. Particularly, the exponential function is not surjective over the reals (hence not inversible) but it is surjective and injective over the positive reals.
– Mefitico
Nov 21 '18 at 18:51
add a comment |
From my understanding the definition of a function's inverse is as follows.
Take a function $f$ which has the inverse $f^{-1}$. This would mean that $f(f^{-1}(x)) = x$ and that $f^{-1}(f(x)) = x$ for every real value for $x$. Right? And this is true for $ln(e^x) = x$ but for $e^{(ln x)} = x$ it only holds true for $x > 0$. Why are they still considered each others' inverse?
real-analysis inverse-function
edited Nov 21 '18 at 17:48
Rócherz
2,7762721
asked Nov 21 '18 at 17:41
CrazyMineCoderCrazyMineCoder
32
From my understanding the definition of a function's inverse is as follows.
Take a function $f$ which has the inverse $f^{-1}$. This would mean that $f(f^{-1}(x)) = x$ and that $f^{-1}(f(x)) = x$ for every real value for $x$. Right? And this is true for $ln(e^x) = x$ but for $e^{(ln x)} = x$ it only holds true for $x > 0$. Why are they still considered each others' inverse?
real-analysis inverse-function
real-analysis inverse-function
edited Nov 21 '18 at 17:48
Rócherz
2,7762721
asked Nov 21 '18 at 17:41
CrazyMineCoderCrazyMineCoder
32
edited Nov 21 '18 at 17:48
Rócherz
2,7762721
asked Nov 21 '18 at 17:41
CrazyMineCoderCrazyMineCoder
32
edited Nov 21 '18 at 17:48
Rócherz
2,7762721
edited Nov 21 '18 at 17:48
Rócherz
2,7762721
edited Nov 21 '18 at 17:48
Rócherz
2,7762721
2,7762721
asked Nov 21 '18 at 17:41
CrazyMineCoderCrazyMineCoder
32
asked Nov 21 '18 at 17:41
CrazyMineCoderCrazyMineCoder
32
asked Nov 21 '18 at 17:41
CrazyMineCoderCrazyMineCoder
32
32
1
It's true because the range of $xmapsto e^x$ for $xin mathbb{R}$ is $(0, infty)$, so of course it only makes sense for the inverse to be defined on $(0, infty)$. Strictly speaking, your definition of inverse isn't quite right; the inverse of a function $f : Xto Y$ is defined $f^{-1} : Yto X$.
– Michael Lee
Nov 21 '18 at 17:46
As @Micheal Lee explained, if f:A→B then it's inverse is f−1:B→A, and these sets are not necessarily the all the real numbers. Particularly, the exponential function is not surjective over the reals (hence not inversible) but it is surjective and injective over the positive reals.
– Mefitico
Nov 21 '18 at 18:51
add a comment |
1
It's true because the range of $xmapsto e^x$ for $xin mathbb{R}$ is $(0, infty)$, so of course it only makes sense for the inverse to be defined on $(0, infty)$. Strictly speaking, your definition of inverse isn't quite right; the inverse of a function $f : Xto Y$ is defined $f^{-1} : Yto X$.
– Michael Lee
Nov 21 '18 at 17:46
As @Micheal Lee explained, if f:A→B then it's inverse is f−1:B→A, and these sets are not necessarily the all the real numbers. Particularly, the exponential function is not surjective over the reals (hence not inversible) but it is surjective and injective over the positive reals.
– Mefitico
Nov 21 '18 at 18:51
1
1
It's true because the range of $xmapsto e^x$ for $xin mathbb{R}$ is $(0, infty)$, so of course it only makes sense for the inverse to be defined on $(0, infty)$. Strictly speaking, your definition of inverse isn't quite right; the inverse of a function $f : Xto Y$ is defined $f^{-1} : Yto X$.
– Michael Lee
Nov 21 '18 at 17:46
It's true because the range of $xmapsto e^x$ for $xin mathbb{R}$ is $(0, infty)$, so of course it only makes sense for the inverse to be defined on $(0, infty)$. Strictly speaking, your definition of inverse isn't quite right; the inverse of a function $f : Xto Y$ is defined $f^{-1} : Yto X$.
– Michael Lee
Nov 21 '18 at 17:46
As @Micheal Lee explained, if f:A→B then it's inverse is f−1:B→A, and these sets are not necessarily the all the real numbers. Particularly, the exponential function is not surjective over the reals (hence not inversible) but it is surjective and injective over the positive reals.
– Mefitico
Nov 21 '18 at 18:51
As @Micheal Lee explained, if f:A→B then it's inverse is f−1:B→A, and these sets are not necessarily the all the real numbers. Particularly, the exponential function is not surjective over the reals (hence not inversible) but it is surjective and injective over the positive reals.
– Mefitico
Nov 21 '18 at 18:51
add a comment |
2 Answers
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By definition, inverse functions have the other one’s domain and range.
The function $f(x) = e^x$ has a domain $x in mathbb{R}$ (all real numbers) and range of $y > 0$ (all positive numbers).
Therefore, $f^{-1}(x) = ln x$ has a domain $x > 0$ and range $y in mathbb{R}$.
$e^{ln x}$ being defined for $x > 0$ has to do with the domain of $ln x$.
$(fcirc f^{-1})(x) = x$ given that $x$ lies within the domain of $f^{-1}(x)$ and that $f^{-1}(x)$ lies within the domain of $f(x)$, which is the case here if $x>0$.
answered Nov 21 '18 at 17:47
KM101KM101
5,5511423
add a comment |
If you have two sets $A,B$ (they may be the same set, or they may not), and two functions $f:Ato B$ and $g:Bto A$, then $f$ and $g$ are said to be eachother's inverses if $g(f(a))=a$ for all $ain A$ and $f(g(b))=b$ for all $bin B$.
In this case, your two sets are the set of real numbers $Bbb R$ and the set of positive real numbers $Bbb R^+$.
answered Nov 21 '18 at 17:47
ArthurArthur
111k7105186
add a comment |
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2 Answers
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2 Answers
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By definition, inverse functions have the other one’s domain and range.
The function $f(x) = e^x$ has a domain $x in mathbb{R}$ (all real numbers) and range of $y > 0$ (all positive numbers).
Therefore, $f^{-1}(x) = ln x$ has a domain $x > 0$ and range $y in mathbb{R}$.
$e^{ln x}$ being defined for $x > 0$ has to do with the domain of $ln x$.
$(fcirc f^{-1})(x) = x$ given that $x$ lies within the domain of $f^{-1}(x)$ and that $f^{-1}(x)$ lies within the domain of $f(x)$, which is the case here if $x>0$.
answered Nov 21 '18 at 17:47
KM101KM101
5,5511423
add a comment |
By definition, inverse functions have the other one’s domain and range.
The function $f(x) = e^x$ has a domain $x in mathbb{R}$ (all real numbers) and range of $y > 0$ (all positive numbers).
Therefore, $f^{-1}(x) = ln x$ has a domain $x > 0$ and range $y in mathbb{R}$.
$e^{ln x}$ being defined for $x > 0$ has to do with the domain of $ln x$.
$(fcirc f^{-1})(x) = x$ given that $x$ lies within the domain of $f^{-1}(x)$ and that $f^{-1}(x)$ lies within the domain of $f(x)$, which is the case here if $x>0$.
answered Nov 21 '18 at 17:47
KM101KM101
5,5511423
add a comment |
By definition, inverse functions have the other one’s domain and range.
The function $f(x) = e^x$ has a domain $x in mathbb{R}$ (all real numbers) and range of $y > 0$ (all positive numbers).
Therefore, $f^{-1}(x) = ln x$ has a domain $x > 0$ and range $y in mathbb{R}$.
$e^{ln x}$ being defined for $x > 0$ has to do with the domain of $ln x$.
$(fcirc f^{-1})(x) = x$ given that $x$ lies within the domain of $f^{-1}(x)$ and that $f^{-1}(x)$ lies within the domain of $f(x)$, which is the case here if $x>0$.
answered Nov 21 '18 at 17:47
KM101KM101
5,5511423
By definition, inverse functions have the other one’s domain and range.
The function $f(x) = e^x$ has a domain $x in mathbb{R}$ (all real numbers) and range of $y > 0$ (all positive numbers).
Therefore, $f^{-1}(x) = ln x$ has a domain $x > 0$ and range $y in mathbb{R}$.
$e^{ln x}$ being defined for $x > 0$ has to do with the domain of $ln x$.
$(fcirc f^{-1})(x) = x$ given that $x$ lies within the domain of $f^{-1}(x)$ and that $f^{-1}(x)$ lies within the domain of $f(x)$, which is the case here if $x>0$.
answered Nov 21 '18 at 17:47
KM101KM101
5,5511423
edited Nov 21 '18 at 18:10
answered Nov 21 '18 at 17:47
KM101KM101
5,5511423
answered Nov 21 '18 at 17:47
KM101KM101
5,5511423
answered Nov 21 '18 at 17:47
KM101KM101
5,5511423
5,5511423
add a comment |
add a comment |
If you have two sets $A,B$ (they may be the same set, or they may not), and two functions $f:Ato B$ and $g:Bto A$, then $f$ and $g$ are said to be eachother's inverses if $g(f(a))=a$ for all $ain A$ and $f(g(b))=b$ for all $bin B$.
In this case, your two sets are the set of real numbers $Bbb R$ and the set of positive real numbers $Bbb R^+$.
answered Nov 21 '18 at 17:47
ArthurArthur
111k7105186
add a comment |
If you have two sets $A,B$ (they may be the same set, or they may not), and two functions $f:Ato B$ and $g:Bto A$, then $f$ and $g$ are said to be eachother's inverses if $g(f(a))=a$ for all $ain A$ and $f(g(b))=b$ for all $bin B$.
In this case, your two sets are the set of real numbers $Bbb R$ and the set of positive real numbers $Bbb R^+$.
answered Nov 21 '18 at 17:47
ArthurArthur
111k7105186
add a comment |
If you have two sets $A,B$ (they may be the same set, or they may not), and two functions $f:Ato B$ and $g:Bto A$, then $f$ and $g$ are said to be eachother's inverses if $g(f(a))=a$ for all $ain A$ and $f(g(b))=b$ for all $bin B$.
In this case, your two sets are the set of real numbers $Bbb R$ and the set of positive real numbers $Bbb R^+$.
answered Nov 21 '18 at 17:47
ArthurArthur
111k7105186
If you have two sets $A,B$ (they may be the same set, or they may not), and two functions $f:Ato B$ and $g:Bto A$, then $f$ and $g$ are said to be eachother's inverses if $g(f(a))=a$ for all $ain A$ and $f(g(b))=b$ for all $bin B$.
In this case, your two sets are the set of real numbers $Bbb R$ and the set of positive real numbers $Bbb R^+$.
answered Nov 21 '18 at 17:47
ArthurArthur
111k7105186
answered Nov 21 '18 at 17:47
ArthurArthur
111k7105186
answered Nov 21 '18 at 17:47
ArthurArthur
111k7105186
answered Nov 21 '18 at 17:47
ArthurArthur
111k7105186
111k7105186
add a comment |
add a comment |
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1
It's true because the range of $xmapsto e^x$ for $xin mathbb{R}$ is $(0, infty)$, so of course it only makes sense for the inverse to be defined on $(0, infty)$. Strictly speaking, your definition of inverse isn't quite right; the inverse of a function $f : Xto Y$ is defined $f^{-1} : Yto X$.
– Michael Lee
Nov 21 '18 at 17:46
As @Micheal Lee explained, if f:A→B then it's inverse is f−1:B→A, and these sets are not necessarily the all the real numbers. Particularly, the exponential function is not surjective over the reals (hence not inversible) but it is surjective and injective over the positive reals.
– Mefitico
Nov 21 '18 at 18:51