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











$endgroup$








  • 5




    $begingroup$
    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.
    10 hours 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
    10 hours ago










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











$endgroup$








  • 5




    $begingroup$
    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.
    10 hours 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
    10 hours ago










  • $begingroup$
    The history is worth considering: Natural log predated Euler's $e^x$
    $endgroup$
    – TurlocTheRed
    32 mins 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 27 mins ago









psmears

70949




70949










asked 10 hours ago









AashishAashish

837




837








  • 5




    $begingroup$
    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.
    10 hours 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
    10 hours ago










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














  • 5




    $begingroup$
    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.
    10 hours 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
    10 hours ago










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








5




5




$begingroup$
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.
10 hours 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.
10 hours 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
10 hours 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
10 hours ago












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




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










5 Answers
5






active

oldest

votes


















11












$begingroup$

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









$endgroup$





















    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











    $endgroup$





















      2












      $begingroup$

      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









      $endgroup$





















        1












        $begingroup$

        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









        $endgroup$





















          0












          $begingroup$

          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
            32 mins ago











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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          11












          $begingroup$

          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









          $endgroup$


















            11












            $begingroup$

            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









            $endgroup$
















              11












              11








              11





              $begingroup$

              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 9 hours ago









              Rhys HughesRhys Hughes

              5,9261529




              5,9261529























                  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 9 hours ago

























                      answered 10 hours ago









                      FredFred

                      44.9k1846




                      44.9k1846























                          2












                          $begingroup$

                          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









                          $endgroup$


















                            2












                            $begingroup$

                            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









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              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 10 hours ago









                              Naman KumarNaman Kumar

                              8410




                              8410























                                  1












                                  $begingroup$

                                  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









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    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









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      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 28 mins ago









                                      AcccumulationAcccumulation

                                      6,8442618




                                      6,8442618























                                          0












                                          $begingroup$

                                          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
                                            32 mins ago
















                                          0












                                          $begingroup$

                                          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
                                            32 mins ago














                                          0












                                          0








                                          0





                                          $begingroup$

                                          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 9 hours ago

























                                          answered 9 hours ago









                                          Yves DaoustYves Daoust

                                          125k671222




                                          125k671222












                                          • $begingroup$
                                            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
                                            32 mins 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
                                            32 mins 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
                                          32 mins 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
                                          32 mins ago


















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