A quick question about logs












3












$begingroup$


$$ln(e^x)=x$$
Is this right?
If yes, then please tell me how.




I know
$e^{ln x}$ $=x$.






calculus logarithms







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
    $endgroup$
    – Matti P.
    2 days ago








  • 3




    $begingroup$
    Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
    $endgroup$
    – maxmilgram
    2 days ago










  • $begingroup$
    The history is worth considering: Natural log predated Euler's $e^x$
    $endgroup$
    – TurlocTheRed
    2 days ago










  • $begingroup$
    Are you assuming that $x$ is a real number?
    $endgroup$
    – Dan
    2 days ago










  • $begingroup$
    It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
    $endgroup$
    – Kemono Chen
    2 days ago
















3












$begingroup$


$$ln(e^x)=x$$
Is this right?
If yes, then please tell me how.




I know
$e^{ln x}$ $=x$.






calculus logarithms







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
    $endgroup$
    – Matti P.
    2 days ago








  • 3




    $begingroup$
    Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
    $endgroup$
    – maxmilgram
    2 days ago










  • $begingroup$
    The history is worth considering: Natural log predated Euler's $e^x$
    $endgroup$
    – TurlocTheRed
    2 days ago










  • $begingroup$
    Are you assuming that $x$ is a real number?
    $endgroup$
    – Dan
    2 days ago










  • $begingroup$
    It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
    $endgroup$
    – Kemono Chen
    2 days ago














3












3








3


0



$begingroup$


$$ln(e^x)=x$$
Is this right?
If yes, then please tell me how.




I know
$e^{ln x}$ $=x$.






calculus logarithms







share|cite|improve this question











$endgroup$




$$ln(e^x)=x$$
Is this right?
If yes, then please tell me how.




I know
$e^{ln x}$ $=x$.






calculus logarithms




calculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









psmears

70949




70949










asked 2 days ago









AashishAashish

888




888








  • 6




    $begingroup$
    It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
    $endgroup$
    – Matti P.
    2 days ago








  • 3




    $begingroup$
    Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
    $endgroup$
    – maxmilgram
    2 days ago










  • $begingroup$
    The history is worth considering: Natural log predated Euler's $e^x$
    $endgroup$
    – TurlocTheRed
    2 days ago










  • $begingroup$
    Are you assuming that $x$ is a real number?
    $endgroup$
    – Dan
    2 days ago










  • $begingroup$
    It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
    $endgroup$
    – Kemono Chen
    2 days ago














  • 6




    $begingroup$
    It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
    $endgroup$
    – Matti P.
    2 days ago








  • 3




    $begingroup$
    Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
    $endgroup$
    – maxmilgram
    2 days ago










  • $begingroup$
    The history is worth considering: Natural log predated Euler's $e^x$
    $endgroup$
    – TurlocTheRed
    2 days ago










  • $begingroup$
    Are you assuming that $x$ is a real number?
    $endgroup$
    – Dan
    2 days ago










  • $begingroup$
    It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
    $endgroup$
    – Kemono Chen
    2 days ago








6




6




$begingroup$
It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
$endgroup$
– Matti P.
2 days ago






$begingroup$
It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
$endgroup$
– Matti P.
2 days ago






3




3




$begingroup$
Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
$endgroup$
– maxmilgram
2 days ago




$begingroup$
Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
$endgroup$
– maxmilgram
2 days ago












$begingroup$
The history is worth considering: Natural log predated Euler's $e^x$
$endgroup$
– TurlocTheRed
2 days ago




$begingroup$
The history is worth considering: Natural log predated Euler's $e^x$
$endgroup$
– TurlocTheRed
2 days ago












$begingroup$
Are you assuming that $x$ is a real number?
$endgroup$
– Dan
2 days ago




$begingroup$
Are you assuming that $x$ is a real number?
$endgroup$
– Dan
2 days ago












$begingroup$
It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
$endgroup$
– Kemono Chen
2 days ago




$begingroup$
It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
$endgroup$
– Kemono Chen
2 days ago










5 Answers
5






active

oldest

votes


















14












$begingroup$

Well:



$$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
By the logarithmic power rule. What you seek comes from this.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Isn't the power rule a stronger property than what is to prove ?
    $endgroup$
    – Yves Daoust
    yesterday










  • $begingroup$
    Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
    $endgroup$
    – Rhys Hughes
    yesterday



















7












$begingroup$

Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.



By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$



Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$






share|cite|improve this answer











$endgroup$





















    5












    $begingroup$

    Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?



    In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.



    In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.



    I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm






    share|cite|improve this answer









    $endgroup$





















      4












      $begingroup$

      With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$



      There is no need to refer to inversion nor any other property than the given.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
        $endgroup$
        – Don Hatch
        2 days ago










      • $begingroup$
        @DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
        $endgroup$
        – Brevan Ellefsen
        yesterday










      • $begingroup$
        @DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
        $endgroup$
        – Yves Daoust
        yesterday





















      4












      $begingroup$

      For real numbers, it's true. But



      $ln e^{2pi i}=ln 1 = 0$



      This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.






      share|cite|improve this answer









      $endgroup$













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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        14












        $begingroup$

        Well:



        $$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
        By the logarithmic power rule. What you seek comes from this.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Isn't the power rule a stronger property than what is to prove ?
          $endgroup$
          – Yves Daoust
          yesterday










        • $begingroup$
          Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
          $endgroup$
          – Rhys Hughes
          yesterday
















        14












        $begingroup$

        Well:



        $$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
        By the logarithmic power rule. What you seek comes from this.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Isn't the power rule a stronger property than what is to prove ?
          $endgroup$
          – Yves Daoust
          yesterday










        • $begingroup$
          Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
          $endgroup$
          – Rhys Hughes
          yesterday














        14












        14








        14





        $begingroup$

        Well:



        $$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
        By the logarithmic power rule. What you seek comes from this.






        share|cite|improve this answer









        $endgroup$



        Well:



        $$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
        By the logarithmic power rule. What you seek comes from this.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Rhys HughesRhys Hughes

        5,7691530




        5,7691530












        • $begingroup$
          Isn't the power rule a stronger property than what is to prove ?
          $endgroup$
          – Yves Daoust
          yesterday










        • $begingroup$
          Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
          $endgroup$
          – Rhys Hughes
          yesterday


















        • $begingroup$
          Isn't the power rule a stronger property than what is to prove ?
          $endgroup$
          – Yves Daoust
          yesterday










        • $begingroup$
          Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
          $endgroup$
          – Rhys Hughes
          yesterday
















        $begingroup$
        Isn't the power rule a stronger property than what is to prove ?
        $endgroup$
        – Yves Daoust
        yesterday




        $begingroup$
        Isn't the power rule a stronger property than what is to prove ?
        $endgroup$
        – Yves Daoust
        yesterday












        $begingroup$
        Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
        $endgroup$
        – Rhys Hughes
        yesterday




        $begingroup$
        Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
        $endgroup$
        – Rhys Hughes
        yesterday











        7












        $begingroup$

        Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.



        By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$



        Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$






        share|cite|improve this answer











        $endgroup$


















          7












          $begingroup$

          Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.



          By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$



          Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$






          share|cite|improve this answer











          $endgroup$
















            7












            7








            7





            $begingroup$

            Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.



            By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$



            Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$






            share|cite|improve this answer











            $endgroup$



            Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.



            By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$



            Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            FredFred

            45.1k1847




            45.1k1847























                5












                $begingroup$

                Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?



                In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.



                In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.



                I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm






                share|cite|improve this answer









                $endgroup$


















                  5












                  $begingroup$

                  Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?



                  In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.



                  In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.



                  I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm






                  share|cite|improve this answer









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?



                    In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.



                    In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.



                    I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm






                    share|cite|improve this answer









                    $endgroup$



                    Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?



                    In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.



                    In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.



                    I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    Naman KumarNaman Kumar

                    12010




                    12010























                        4












                        $begingroup$

                        With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$



                        There is no need to refer to inversion nor any other property than the given.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
                          $endgroup$
                          – Don Hatch
                          2 days ago










                        • $begingroup$
                          @DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
                          $endgroup$
                          – Brevan Ellefsen
                          yesterday










                        • $begingroup$
                          @DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
                          $endgroup$
                          – Yves Daoust
                          yesterday


















                        4












                        $begingroup$

                        With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$



                        There is no need to refer to inversion nor any other property than the given.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
                          $endgroup$
                          – Don Hatch
                          2 days ago










                        • $begingroup$
                          @DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
                          $endgroup$
                          – Brevan Ellefsen
                          yesterday










                        • $begingroup$
                          @DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
                          $endgroup$
                          – Yves Daoust
                          yesterday
















                        4












                        4








                        4





                        $begingroup$

                        With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$



                        There is no need to refer to inversion nor any other property than the given.






                        share|cite|improve this answer











                        $endgroup$



                        With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$



                        There is no need to refer to inversion nor any other property than the given.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 2 days ago

























                        answered 2 days ago









                        Yves DaoustYves Daoust

                        125k671223




                        125k671223












                        • $begingroup$
                          Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
                          $endgroup$
                          – Don Hatch
                          2 days ago










                        • $begingroup$
                          @DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
                          $endgroup$
                          – Brevan Ellefsen
                          yesterday










                        • $begingroup$
                          @DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
                          $endgroup$
                          – Yves Daoust
                          yesterday




















                        • $begingroup$
                          Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
                          $endgroup$
                          – Don Hatch
                          2 days ago










                        • $begingroup$
                          @DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
                          $endgroup$
                          – Brevan Ellefsen
                          yesterday










                        • $begingroup$
                          @DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
                          $endgroup$
                          – Yves Daoust
                          yesterday


















                        $begingroup$
                        Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
                        $endgroup$
                        – Don Hatch
                        2 days ago




                        $begingroup$
                        Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
                        $endgroup$
                        – Don Hatch
                        2 days ago












                        $begingroup$
                        @DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
                        $endgroup$
                        – Brevan Ellefsen
                        yesterday




                        $begingroup$
                        @DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
                        $endgroup$
                        – Brevan Ellefsen
                        yesterday












                        $begingroup$
                        @DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
                        $endgroup$
                        – Yves Daoust
                        yesterday






                        $begingroup$
                        @DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
                        $endgroup$
                        – Yves Daoust
                        yesterday













                        4












                        $begingroup$

                        For real numbers, it's true. But



                        $ln e^{2pi i}=ln 1 = 0$



                        This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.






                        share|cite|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          For real numbers, it's true. But



                          $ln e^{2pi i}=ln 1 = 0$



                          This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.






                          share|cite|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            For real numbers, it's true. But



                            $ln e^{2pi i}=ln 1 = 0$



                            This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.






                            share|cite|improve this answer









                            $endgroup$



                            For real numbers, it's true. But



                            $ln e^{2pi i}=ln 1 = 0$



                            This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            AcccumulationAcccumulation

                            6,9142618




                            6,9142618






























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