Is there a logarithm base for which the logarithm becomes an identity function?












2












$begingroup$


Is there a base $b$ such that:



$$log_b x = x $$



(The only one that comes to mind would be the invalid case of $log_1 1 = 1 $.)



I'm fairly certain the answer is no, but I can't find a clear justification for it.



(I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)










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  • $begingroup$
    Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
    $endgroup$
    – Somos
    48 mins ago












  • $begingroup$
    The question body is not an appropriate place for any answer. If you want to summarize the other answers, do it as an answer. Make it community wiki if you don't want to feel like stealing credit.
    $endgroup$
    – Nij
    25 mins ago










  • $begingroup$
    Good point, I didn't know I could answer my own question. I'll fix it, the accepted answer shouldn't lose visibility with this text added a posteriori. I don't feel however as stealing credit from an answer I've accepted and whose author I'm mentioning at the beginning of the text.
    $endgroup$
    – schomatis
    19 mins ago
















2












$begingroup$


Is there a base $b$ such that:



$$log_b x = x $$



(The only one that comes to mind would be the invalid case of $log_1 1 = 1 $.)



I'm fairly certain the answer is no, but I can't find a clear justification for it.



(I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)










share|cite|improve this question









New contributor




schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
    $endgroup$
    – Somos
    48 mins ago












  • $begingroup$
    The question body is not an appropriate place for any answer. If you want to summarize the other answers, do it as an answer. Make it community wiki if you don't want to feel like stealing credit.
    $endgroup$
    – Nij
    25 mins ago










  • $begingroup$
    Good point, I didn't know I could answer my own question. I'll fix it, the accepted answer shouldn't lose visibility with this text added a posteriori. I don't feel however as stealing credit from an answer I've accepted and whose author I'm mentioning at the beginning of the text.
    $endgroup$
    – schomatis
    19 mins ago














2












2








2





$begingroup$


Is there a base $b$ such that:



$$log_b x = x $$



(The only one that comes to mind would be the invalid case of $log_1 1 = 1 $.)



I'm fairly certain the answer is no, but I can't find a clear justification for it.



(I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)










share|cite|improve this question









New contributor




schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Is there a base $b$ such that:



$$log_b x = x $$



(The only one that comes to mind would be the invalid case of $log_1 1 = 1 $.)



I'm fairly certain the answer is no, but I can't find a clear justification for it.



(I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)







logarithms






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schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question




share|cite|improve this question








edited 17 mins ago







schomatis













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asked 5 hours ago









schomatisschomatis

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163




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schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
    $endgroup$
    – Somos
    48 mins ago












  • $begingroup$
    The question body is not an appropriate place for any answer. If you want to summarize the other answers, do it as an answer. Make it community wiki if you don't want to feel like stealing credit.
    $endgroup$
    – Nij
    25 mins ago










  • $begingroup$
    Good point, I didn't know I could answer my own question. I'll fix it, the accepted answer shouldn't lose visibility with this text added a posteriori. I don't feel however as stealing credit from an answer I've accepted and whose author I'm mentioning at the beginning of the text.
    $endgroup$
    – schomatis
    19 mins ago


















  • $begingroup$
    Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
    $endgroup$
    – Somos
    48 mins ago












  • $begingroup$
    The question body is not an appropriate place for any answer. If you want to summarize the other answers, do it as an answer. Make it community wiki if you don't want to feel like stealing credit.
    $endgroup$
    – Nij
    25 mins ago










  • $begingroup$
    Good point, I didn't know I could answer my own question. I'll fix it, the accepted answer shouldn't lose visibility with this text added a posteriori. I don't feel however as stealing credit from an answer I've accepted and whose author I'm mentioning at the beginning of the text.
    $endgroup$
    – schomatis
    19 mins ago
















$begingroup$
Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
$endgroup$
– Somos
48 mins ago






$begingroup$
Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
$endgroup$
– Somos
48 mins ago














$begingroup$
The question body is not an appropriate place for any answer. If you want to summarize the other answers, do it as an answer. Make it community wiki if you don't want to feel like stealing credit.
$endgroup$
– Nij
25 mins ago




$begingroup$
The question body is not an appropriate place for any answer. If you want to summarize the other answers, do it as an answer. Make it community wiki if you don't want to feel like stealing credit.
$endgroup$
– Nij
25 mins ago












$begingroup$
Good point, I didn't know I could answer my own question. I'll fix it, the accepted answer shouldn't lose visibility with this text added a posteriori. I don't feel however as stealing credit from an answer I've accepted and whose author I'm mentioning at the beginning of the text.
$endgroup$
– schomatis
19 mins ago




$begingroup$
Good point, I didn't know I could answer my own question. I'll fix it, the accepted answer shouldn't lose visibility with this text added a posteriori. I don't feel however as stealing credit from an answer I've accepted and whose author I'm mentioning at the beginning of the text.
$endgroup$
– schomatis
19 mins ago










7 Answers
7






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oldest

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8












$begingroup$

For a function to be a logarithm, it should satisfy the law of logarithms:
$log ab = log a + log b$, for $a,b gt 0$.
If it were the identity function, this would become $ab = a + b$, which clearly is not always true.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Note that
    $$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
    Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.



    But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
    $$log_{sqrt{2}}2=2.$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      In general
      $$log_b a=c$$
      is the same as
      $$b^c=a$$
      so you can leave logs behind and focus on solutions to
      $$b^x=x$$






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        If $b^k = k$ for all $k$ then



        $(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.



        ....



        Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.



        Likewise $log_b b = 1$ and presumably $b ne 1$






        share|cite|improve this answer











        $endgroup$





















          1












          $begingroup$

          No, it can't. For any base $b$, there is some real constant $C$, s.t.
          $$
          log_b x = C ln x
          $$

          If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              And why is $y=log_b x$ not a straight line?
              $endgroup$
              – Henning Makholm
              5 hours ago










            • $begingroup$
              @HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
              $endgroup$
              – Vasya
              5 hours ago










            • $begingroup$
              $frac0x$ is a constant function on a useful subset of $mathbb R$.
              $endgroup$
              – Henning Makholm
              5 hours ago












            • $begingroup$
              @HenningMakholm: we both know that $c ne 0$ in this case, I was trying to explain it at the simplest level for the original poster.
              $endgroup$
              – Vasya
              35 mins ago



















            0












            $begingroup$

            I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!



            First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:



            $$ b^x=x $$



            In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).



            An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$



            $$ b^{x+1} = x + 1 iff b^x * b = x + 1 $$



            so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).



            I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).



            As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.






            share|cite|improve this answer








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              7 Answers
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              active

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              7 Answers
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              active

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              active

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              active

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              8












              $begingroup$

              For a function to be a logarithm, it should satisfy the law of logarithms:
              $log ab = log a + log b$, for $a,b gt 0$.
              If it were the identity function, this would become $ab = a + b$, which clearly is not always true.






              share|cite|improve this answer









              $endgroup$


















                8












                $begingroup$

                For a function to be a logarithm, it should satisfy the law of logarithms:
                $log ab = log a + log b$, for $a,b gt 0$.
                If it were the identity function, this would become $ab = a + b$, which clearly is not always true.






                share|cite|improve this answer









                $endgroup$
















                  8












                  8








                  8





                  $begingroup$

                  For a function to be a logarithm, it should satisfy the law of logarithms:
                  $log ab = log a + log b$, for $a,b gt 0$.
                  If it were the identity function, this would become $ab = a + b$, which clearly is not always true.






                  share|cite|improve this answer









                  $endgroup$



                  For a function to be a logarithm, it should satisfy the law of logarithms:
                  $log ab = log a + log b$, for $a,b gt 0$.
                  If it were the identity function, this would become $ab = a + b$, which clearly is not always true.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 5 hours ago









                  FredHFredH

                  1,258612




                  1,258612























                      4












                      $begingroup$

                      Note that
                      $$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
                      Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.



                      But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
                      $$log_{sqrt{2}}2=2.$$






                      share|cite|improve this answer









                      $endgroup$


















                        4












                        $begingroup$

                        Note that
                        $$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
                        Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.



                        But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
                        $$log_{sqrt{2}}2=2.$$






                        share|cite|improve this answer









                        $endgroup$
















                          4












                          4








                          4





                          $begingroup$

                          Note that
                          $$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
                          Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.



                          But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
                          $$log_{sqrt{2}}2=2.$$






                          share|cite|improve this answer









                          $endgroup$



                          Note that
                          $$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
                          Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.



                          But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
                          $$log_{sqrt{2}}2=2.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 5 hours ago









                          Eclipse SunEclipse Sun

                          7,7101437




                          7,7101437























                              2












                              $begingroup$

                              In general
                              $$log_b a=c$$
                              is the same as
                              $$b^c=a$$
                              so you can leave logs behind and focus on solutions to
                              $$b^x=x$$






                              share|cite|improve this answer









                              $endgroup$


















                                2












                                $begingroup$

                                In general
                                $$log_b a=c$$
                                is the same as
                                $$b^c=a$$
                                so you can leave logs behind and focus on solutions to
                                $$b^x=x$$






                                share|cite|improve this answer









                                $endgroup$
















                                  2












                                  2








                                  2





                                  $begingroup$

                                  In general
                                  $$log_b a=c$$
                                  is the same as
                                  $$b^c=a$$
                                  so you can leave logs behind and focus on solutions to
                                  $$b^x=x$$






                                  share|cite|improve this answer









                                  $endgroup$



                                  In general
                                  $$log_b a=c$$
                                  is the same as
                                  $$b^c=a$$
                                  so you can leave logs behind and focus on solutions to
                                  $$b^x=x$$







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered 5 hours ago









                                  Martin HansenMartin Hansen

                                  18113




                                  18113























                                      1












                                      $begingroup$

                                      If $b^k = k$ for all $k$ then



                                      $(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.



                                      ....



                                      Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.



                                      Likewise $log_b b = 1$ and presumably $b ne 1$






                                      share|cite|improve this answer











                                      $endgroup$


















                                        1












                                        $begingroup$

                                        If $b^k = k$ for all $k$ then



                                        $(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.



                                        ....



                                        Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.



                                        Likewise $log_b b = 1$ and presumably $b ne 1$






                                        share|cite|improve this answer











                                        $endgroup$
















                                          1












                                          1








                                          1





                                          $begingroup$

                                          If $b^k = k$ for all $k$ then



                                          $(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.



                                          ....



                                          Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.



                                          Likewise $log_b b = 1$ and presumably $b ne 1$






                                          share|cite|improve this answer











                                          $endgroup$



                                          If $b^k = k$ for all $k$ then



                                          $(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.



                                          ....



                                          Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.



                                          Likewise $log_b b = 1$ and presumably $b ne 1$







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited 4 hours ago

























                                          answered 5 hours ago









                                          fleabloodfleablood

                                          72k22687




                                          72k22687























                                              1












                                              $begingroup$

                                              No, it can't. For any base $b$, there is some real constant $C$, s.t.
                                              $$
                                              log_b x = C ln x
                                              $$

                                              If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.






                                              share|cite|improve this answer











                                              $endgroup$


















                                                1












                                                $begingroup$

                                                No, it can't. For any base $b$, there is some real constant $C$, s.t.
                                                $$
                                                log_b x = C ln x
                                                $$

                                                If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.






                                                share|cite|improve this answer











                                                $endgroup$
















                                                  1












                                                  1








                                                  1





                                                  $begingroup$

                                                  No, it can't. For any base $b$, there is some real constant $C$, s.t.
                                                  $$
                                                  log_b x = C ln x
                                                  $$

                                                  If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.






                                                  share|cite|improve this answer











                                                  $endgroup$



                                                  No, it can't. For any base $b$, there is some real constant $C$, s.t.
                                                  $$
                                                  log_b x = C ln x
                                                  $$

                                                  If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.







                                                  share|cite|improve this answer














                                                  share|cite|improve this answer



                                                  share|cite|improve this answer








                                                  edited 4 hours ago

























                                                  answered 4 hours ago









                                                  enedilenedil

                                                  1,359620




                                                  1,359620























                                                      0












                                                      $begingroup$

                                                      Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.






                                                      share|cite|improve this answer









                                                      $endgroup$









                                                      • 1




                                                        $begingroup$
                                                        And why is $y=log_b x$ not a straight line?
                                                        $endgroup$
                                                        – Henning Makholm
                                                        5 hours ago










                                                      • $begingroup$
                                                        @HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
                                                        $endgroup$
                                                        – Vasya
                                                        5 hours ago










                                                      • $begingroup$
                                                        $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                        $endgroup$
                                                        – Henning Makholm
                                                        5 hours ago












                                                      • $begingroup$
                                                        @HenningMakholm: we both know that $c ne 0$ in this case, I was trying to explain it at the simplest level for the original poster.
                                                        $endgroup$
                                                        – Vasya
                                                        35 mins ago
















                                                      0












                                                      $begingroup$

                                                      Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.






                                                      share|cite|improve this answer









                                                      $endgroup$









                                                      • 1




                                                        $begingroup$
                                                        And why is $y=log_b x$ not a straight line?
                                                        $endgroup$
                                                        – Henning Makholm
                                                        5 hours ago










                                                      • $begingroup$
                                                        @HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
                                                        $endgroup$
                                                        – Vasya
                                                        5 hours ago










                                                      • $begingroup$
                                                        $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                        $endgroup$
                                                        – Henning Makholm
                                                        5 hours ago












                                                      • $begingroup$
                                                        @HenningMakholm: we both know that $c ne 0$ in this case, I was trying to explain it at the simplest level for the original poster.
                                                        $endgroup$
                                                        – Vasya
                                                        35 mins ago














                                                      0












                                                      0








                                                      0





                                                      $begingroup$

                                                      Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.






                                                      share|cite|improve this answer









                                                      $endgroup$



                                                      Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.







                                                      share|cite|improve this answer












                                                      share|cite|improve this answer



                                                      share|cite|improve this answer










                                                      answered 5 hours ago









                                                      VasyaVasya

                                                      3,9531618




                                                      3,9531618








                                                      • 1




                                                        $begingroup$
                                                        And why is $y=log_b x$ not a straight line?
                                                        $endgroup$
                                                        – Henning Makholm
                                                        5 hours ago










                                                      • $begingroup$
                                                        @HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
                                                        $endgroup$
                                                        – Vasya
                                                        5 hours ago










                                                      • $begingroup$
                                                        $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                        $endgroup$
                                                        – Henning Makholm
                                                        5 hours ago












                                                      • $begingroup$
                                                        @HenningMakholm: we both know that $c ne 0$ in this case, I was trying to explain it at the simplest level for the original poster.
                                                        $endgroup$
                                                        – Vasya
                                                        35 mins ago














                                                      • 1




                                                        $begingroup$
                                                        And why is $y=log_b x$ not a straight line?
                                                        $endgroup$
                                                        – Henning Makholm
                                                        5 hours ago










                                                      • $begingroup$
                                                        @HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
                                                        $endgroup$
                                                        – Vasya
                                                        5 hours ago










                                                      • $begingroup$
                                                        $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                        $endgroup$
                                                        – Henning Makholm
                                                        5 hours ago












                                                      • $begingroup$
                                                        @HenningMakholm: we both know that $c ne 0$ in this case, I was trying to explain it at the simplest level for the original poster.
                                                        $endgroup$
                                                        – Vasya
                                                        35 mins ago








                                                      1




                                                      1




                                                      $begingroup$
                                                      And why is $y=log_b x$ not a straight line?
                                                      $endgroup$
                                                      – Henning Makholm
                                                      5 hours ago




                                                      $begingroup$
                                                      And why is $y=log_b x$ not a straight line?
                                                      $endgroup$
                                                      – Henning Makholm
                                                      5 hours ago












                                                      $begingroup$
                                                      @HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
                                                      $endgroup$
                                                      – Vasya
                                                      5 hours ago




                                                      $begingroup$
                                                      @HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
                                                      $endgroup$
                                                      – Vasya
                                                      5 hours ago












                                                      $begingroup$
                                                      $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      5 hours ago






                                                      $begingroup$
                                                      $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      5 hours ago














                                                      $begingroup$
                                                      @HenningMakholm: we both know that $c ne 0$ in this case, I was trying to explain it at the simplest level for the original poster.
                                                      $endgroup$
                                                      – Vasya
                                                      35 mins ago




                                                      $begingroup$
                                                      @HenningMakholm: we both know that $c ne 0$ in this case, I was trying to explain it at the simplest level for the original poster.
                                                      $endgroup$
                                                      – Vasya
                                                      35 mins ago











                                                      0












                                                      $begingroup$

                                                      I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!



                                                      First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:



                                                      $$ b^x=x $$



                                                      In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).



                                                      An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$



                                                      $$ b^{x+1} = x + 1 iff b^x * b = x + 1 $$



                                                      so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).



                                                      I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).



                                                      As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.






                                                      share|cite|improve this answer








                                                      New contributor




                                                      schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                                      $endgroup$


















                                                        0












                                                        $begingroup$

                                                        I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!



                                                        First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:



                                                        $$ b^x=x $$



                                                        In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).



                                                        An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$



                                                        $$ b^{x+1} = x + 1 iff b^x * b = x + 1 $$



                                                        so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).



                                                        I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).



                                                        As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.






                                                        share|cite|improve this answer








                                                        New contributor




                                                        schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                        Check out our Code of Conduct.






                                                        $endgroup$
















                                                          0












                                                          0








                                                          0





                                                          $begingroup$

                                                          I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!



                                                          First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:



                                                          $$ b^x=x $$



                                                          In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).



                                                          An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$



                                                          $$ b^{x+1} = x + 1 iff b^x * b = x + 1 $$



                                                          so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).



                                                          I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).



                                                          As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.






                                                          share|cite|improve this answer








                                                          New contributor




                                                          schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                          Check out our Code of Conduct.






                                                          $endgroup$



                                                          I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!



                                                          First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:



                                                          $$ b^x=x $$



                                                          In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).



                                                          An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$



                                                          $$ b^{x+1} = x + 1 iff b^x * b = x + 1 $$



                                                          so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).



                                                          I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).



                                                          As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.







                                                          share|cite|improve this answer








                                                          New contributor




                                                          schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                          Check out our Code of Conduct.









                                                          share|cite|improve this answer



                                                          share|cite|improve this answer






                                                          New contributor




                                                          schomatis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                                          answered 17 mins ago









                                                          schomatisschomatis

                                                          163




                                                          163




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