Why are $ln x$ and $e^x$ considered to be each others' inverses?
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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 at 17:48
Rócherz
2,7162721
asked Nov 21 at 17:41
CrazyMineCoder
32
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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 at 17:48
Rócherz
2,7162721
asked Nov 21 at 17:41
CrazyMineCoder
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 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 at 18:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 17:48
Rócherz
2,7162721
asked Nov 21 at 17:41
CrazyMineCoder
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 at 17:48
Rócherz
2,7162721
asked Nov 21 at 17:41
CrazyMineCoder
32
edited Nov 21 at 17:48
Rócherz
2,7162721
asked Nov 21 at 17:41
CrazyMineCoder
32
edited Nov 21 at 17:48
Rócherz
2,7162721
edited Nov 21 at 17:48
Rócherz
2,7162721
edited Nov 21 at 17:48
Rócherz
2,7162721
2,7162721
asked Nov 21 at 17:41
CrazyMineCoder
32
asked Nov 21 at 17:41
CrazyMineCoder
32
asked Nov 21 at 17:41
CrazyMineCoder
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 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 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 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 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 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 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 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 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 at 17:47
KM101
3,609417
add a comment |
up vote
1
down vote
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 at 17:47
Arthur
110k7104186
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 at 17:47
KM101
3,609417
add a comment |
up vote
1
down vote
accepted
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 at 17:47
KM101
3,609417
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 at 17:47
KM101
3,609417
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 at 17:47
KM101
3,609417
edited Nov 21 at 18:10
answered Nov 21 at 17:47
KM101
3,609417
answered Nov 21 at 17:47
KM101
3,609417
answered Nov 21 at 17:47
KM101
3,609417
3,609417
add a comment |
add a comment |
up vote
1
down vote
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 at 17:47
Arthur
110k7104186
add a comment |
up vote
1
down vote
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 at 17:47
Arthur
110k7104186
add a comment |
up vote
1
down vote
up vote
1
down vote
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 at 17:47
Arthur
110k7104186
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 at 17:47
Arthur
110k7104186
answered Nov 21 at 17:47
Arthur
110k7104186
answered Nov 21 at 17:47
Arthur
110k7104186
answered Nov 21 at 17:47
Arthur
110k7104186
110k7104186
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 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 at 18:51