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?










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    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















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?










share|cite|improve this question




















  • 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













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?










share|cite|improve this question















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






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edited Nov 21 at 17:48









Rócherz

2,7162721




2,7162721










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














  • 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










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 $x>0$.






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    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 $Bbb R^+$.






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      2 Answers
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      active

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      2 Answers
      2






      active

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      active

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      active

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      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 $x>0$.






      share|cite|improve this answer



























        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 $x>0$.






        share|cite|improve this answer

























          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 $x>0$.






          share|cite|improve this answer














          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 $x>0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 at 18:10

























          answered Nov 21 at 17:47









          KM101

          3,609417




          3,609417






















              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 $Bbb R^+$.






              share|cite|improve this answer

























                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 $Bbb R^+$.






                share|cite|improve this answer























                  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 $Bbb R^+$.






                  share|cite|improve this answer












                  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 $Bbb R^+$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 at 17:47









                  Arthur

                  110k7104186




                  110k7104186






























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