Is there a logarithm base for which the logarithm becomes an identity function?Why must the base of a logarithm be a positive real number not equal to 1?How to determine periodicity of complex log in different bases?Problems simplifying logarithmic expressionsLogarithm base transformationIs the natural logarithm actually unique as a multiplier?Integral of $b[x]^l[x] = p[x]$Intuition behind logarithm change of baseThe base of a logarithmUnderlying Reason For Taking Log Base 10Logarithm IdentityGeometric interpretation of the Logarithm (in $mathbbR$)

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


Why must the base of a logarithm be a positive real number not equal to 1?How to determine periodicity of complex log in different bases?Problems simplifying logarithmic expressionsLogarithm base transformationIs the natural logarithm actually unique as a multiplier?Integral of $b[x]^l[x] = p[x]$Intuition behind logarithm change of baseThe base of a logarithmUnderlying Reason For Taking Log Base 10Logarithm IdentityGeometric interpretation of the Logarithm (in $mathbbR$)













12












$begingroup$


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











$endgroup$







  • 2




    $begingroup$
    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
    Mar 9 at 3:33










  • $begingroup$
    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
    Mar 9 at 3:39










  • $begingroup$
    Changing the base of the logarithm just scales it.
    $endgroup$
    – user76284
    Mar 10 at 2:28















12












$begingroup$


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











$endgroup$







  • 2




    $begingroup$
    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
    Mar 9 at 3:33










  • $begingroup$
    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
    Mar 9 at 3:39










  • $begingroup$
    Changing the base of the logarithm just scales it.
    $endgroup$
    – user76284
    Mar 10 at 2:28













12












12








12


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











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 9 at 3:41







schomatis

















asked Mar 8 at 21:58









schomatisschomatis

1088




1088







  • 2




    $begingroup$
    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
    Mar 9 at 3:33










  • $begingroup$
    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
    Mar 9 at 3:39










  • $begingroup$
    Changing the base of the logarithm just scales it.
    $endgroup$
    – user76284
    Mar 10 at 2:28












  • 2




    $begingroup$
    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
    Mar 9 at 3:33










  • $begingroup$
    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
    Mar 9 at 3:39










  • $begingroup$
    Changing the base of the logarithm just scales it.
    $endgroup$
    – user76284
    Mar 10 at 2:28







2




2




$begingroup$
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
Mar 9 at 3:33




$begingroup$
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
Mar 9 at 3:33












$begingroup$
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
Mar 9 at 3:39




$begingroup$
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
Mar 9 at 3:39












$begingroup$
Changing the base of the logarithm just scales it.
$endgroup$
– user76284
Mar 10 at 2:28




$begingroup$
Changing the base of the logarithm just scales it.
$endgroup$
– user76284
Mar 10 at 2:28










7 Answers
7






active

oldest

votes


















47












$begingroup$

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









$endgroup$




















    18












    $begingroup$

    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 $b=sqrt2$ and we have
    $$log_sqrt22=2.$$






    share|cite|improve this answer









    $endgroup$




















      8












      $begingroup$

      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 $x/C$, which is clearly false.






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        If log_b X = C ln x and log_b is the identity function, then log_b x = x = C ln x ==> ln x = x/C .
        $endgroup$
        – Adam
        Mar 9 at 16:05







      • 1




        $begingroup$
        Further, this shows that the identity function is not even any sort of limit of logarithm functions.
        $endgroup$
        – R..
        Mar 9 at 22:26


















      7












      $begingroup$

      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











      $endgroup$




















        5












        $begingroup$

        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









        $endgroup$




















          3












          $begingroup$

          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









          $endgroup$




















            1












            $begingroup$

            Logically, $y=x$ is a straight line, $y=log_b x$ is not (otherwise why would we call it non-linear function) so they cannot coincide for all $x$.



            Suppose $y=log_b x$ is a straight line the same as $y=x$ then for any two $x_1$ and $x_2$ such as $x_2=x_1+1$ we should have for the slope: $1=fraclog_b x_2-log_b x_1x_2-x_1=log_b x_2-log_b x_1=log_b fracx_2x_1 rightarrow fracx_2x_1=b$ which leads to a contradiction ($frac43=b, frac54=b$).






            share|cite|improve this answer











            $endgroup$








            • 2




              $begingroup$
              And why is $y=log_b x$ not a straight line?
              $endgroup$
              – Henning Makholm
              Mar 8 at 22:26






            • 1




              $begingroup$
              @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
              Mar 8 at 22:32










            • $begingroup$
              $frac0x$ is a constant function on a useful subset of $mathbb R$.
              $endgroup$
              – Henning Makholm
              Mar 8 at 22:37







            • 2




              $begingroup$
              @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
              Mar 9 at 3:23






            • 2




              $begingroup$
              The OP was asking for an argument, not for a "simplest level" that amounts to an unsupported assertion of the same thing he wanted a proof of.
              $endgroup$
              – Henning Makholm
              Mar 9 at 10:14












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            7 Answers
            7






            active

            oldest

            votes








            7 Answers
            7






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            47












            $begingroup$

            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









            $endgroup$

















              47












              $begingroup$

              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









              $endgroup$















                47












                47








                47





                $begingroup$

                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 Mar 8 at 22:08









                FredHFredH

                3,7051023




                3,7051023





















                    18












                    $begingroup$

                    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 $b=sqrt2$ and we have
                    $$log_sqrt22=2.$$






                    share|cite|improve this answer









                    $endgroup$

















                      18












                      $begingroup$

                      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 $b=sqrt2$ and we have
                      $$log_sqrt22=2.$$






                      share|cite|improve this answer









                      $endgroup$















                        18












                        18








                        18





                        $begingroup$

                        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 $b=sqrt2$ and we have
                        $$log_sqrt22=2.$$






                        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 $b=sqrt2$ and we have
                        $$log_sqrt22=2.$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Mar 8 at 22:06









                        Eclipse SunEclipse Sun

                        8,0051438




                        8,0051438





















                            8












                            $begingroup$

                            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 $x/C$, which is clearly false.






                            share|cite|improve this answer











                            $endgroup$












                            • $begingroup$
                              If log_b X = C ln x and log_b is the identity function, then log_b x = x = C ln x ==> ln x = x/C .
                              $endgroup$
                              – Adam
                              Mar 9 at 16:05







                            • 1




                              $begingroup$
                              Further, this shows that the identity function is not even any sort of limit of logarithm functions.
                              $endgroup$
                              – R..
                              Mar 9 at 22:26















                            8












                            $begingroup$

                            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 $x/C$, which is clearly false.






                            share|cite|improve this answer











                            $endgroup$












                            • $begingroup$
                              If log_b X = C ln x and log_b is the identity function, then log_b x = x = C ln x ==> ln x = x/C .
                              $endgroup$
                              – Adam
                              Mar 9 at 16:05







                            • 1




                              $begingroup$
                              Further, this shows that the identity function is not even any sort of limit of logarithm functions.
                              $endgroup$
                              – R..
                              Mar 9 at 22:26













                            8












                            8








                            8





                            $begingroup$

                            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 $x/C$, 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 $x/C$, which is clearly false.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Mar 9 at 16:15

























                            answered Mar 8 at 23:17









                            enedilenedil

                            1,479720




                            1,479720











                            • $begingroup$
                              If log_b X = C ln x and log_b is the identity function, then log_b x = x = C ln x ==> ln x = x/C .
                              $endgroup$
                              – Adam
                              Mar 9 at 16:05







                            • 1




                              $begingroup$
                              Further, this shows that the identity function is not even any sort of limit of logarithm functions.
                              $endgroup$
                              – R..
                              Mar 9 at 22:26
















                            • $begingroup$
                              If log_b X = C ln x and log_b is the identity function, then log_b x = x = C ln x ==> ln x = x/C .
                              $endgroup$
                              – Adam
                              Mar 9 at 16:05







                            • 1




                              $begingroup$
                              Further, this shows that the identity function is not even any sort of limit of logarithm functions.
                              $endgroup$
                              – R..
                              Mar 9 at 22:26















                            $begingroup$
                            If log_b X = C ln x and log_b is the identity function, then log_b x = x = C ln x ==> ln x = x/C .
                            $endgroup$
                            – Adam
                            Mar 9 at 16:05





                            $begingroup$
                            If log_b X = C ln x and log_b is the identity function, then log_b x = x = C ln x ==> ln x = x/C .
                            $endgroup$
                            – Adam
                            Mar 9 at 16:05





                            1




                            1




                            $begingroup$
                            Further, this shows that the identity function is not even any sort of limit of logarithm functions.
                            $endgroup$
                            – R..
                            Mar 9 at 22:26




                            $begingroup$
                            Further, this shows that the identity function is not even any sort of limit of logarithm functions.
                            $endgroup$
                            – R..
                            Mar 9 at 22:26











                            7












                            $begingroup$

                            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











                            $endgroup$

















                              7












                              $begingroup$

                              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











                              $endgroup$















                                7












                                7








                                7





                                $begingroup$

                                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 Mar 8 at 23:23

























                                answered Mar 8 at 22:07









                                fleabloodfleablood

                                73.8k22891




                                73.8k22891





















                                    5












                                    $begingroup$

                                    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









                                    $endgroup$

















                                      5












                                      $begingroup$

                                      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









                                      $endgroup$















                                        5












                                        5








                                        5





                                        $begingroup$

                                        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 Mar 8 at 22:05









                                        Martin HansenMartin Hansen

                                        780114




                                        780114





















                                            3












                                            $begingroup$

                                            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









                                            $endgroup$

















                                              3












                                              $begingroup$

                                              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









                                              $endgroup$















                                                3












                                                3








                                                3





                                                $begingroup$

                                                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









                                                $endgroup$



                                                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












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Mar 9 at 3:40









                                                schomatisschomatis

                                                1088




                                                1088





















                                                    1












                                                    $begingroup$

                                                    Logically, $y=x$ is a straight line, $y=log_b x$ is not (otherwise why would we call it non-linear function) so they cannot coincide for all $x$.



                                                    Suppose $y=log_b x$ is a straight line the same as $y=x$ then for any two $x_1$ and $x_2$ such as $x_2=x_1+1$ we should have for the slope: $1=fraclog_b x_2-log_b x_1x_2-x_1=log_b x_2-log_b x_1=log_b fracx_2x_1 rightarrow fracx_2x_1=b$ which leads to a contradiction ($frac43=b, frac54=b$).






                                                    share|cite|improve this answer











                                                    $endgroup$








                                                    • 2




                                                      $begingroup$
                                                      And why is $y=log_b x$ not a straight line?
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 8 at 22:26






                                                    • 1




                                                      $begingroup$
                                                      @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
                                                      Mar 8 at 22:32










                                                    • $begingroup$
                                                      $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 8 at 22:37







                                                    • 2




                                                      $begingroup$
                                                      @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
                                                      Mar 9 at 3:23






                                                    • 2




                                                      $begingroup$
                                                      The OP was asking for an argument, not for a "simplest level" that amounts to an unsupported assertion of the same thing he wanted a proof of.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 9 at 10:14
















                                                    1












                                                    $begingroup$

                                                    Logically, $y=x$ is a straight line, $y=log_b x$ is not (otherwise why would we call it non-linear function) so they cannot coincide for all $x$.



                                                    Suppose $y=log_b x$ is a straight line the same as $y=x$ then for any two $x_1$ and $x_2$ such as $x_2=x_1+1$ we should have for the slope: $1=fraclog_b x_2-log_b x_1x_2-x_1=log_b x_2-log_b x_1=log_b fracx_2x_1 rightarrow fracx_2x_1=b$ which leads to a contradiction ($frac43=b, frac54=b$).






                                                    share|cite|improve this answer











                                                    $endgroup$








                                                    • 2




                                                      $begingroup$
                                                      And why is $y=log_b x$ not a straight line?
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 8 at 22:26






                                                    • 1




                                                      $begingroup$
                                                      @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
                                                      Mar 8 at 22:32










                                                    • $begingroup$
                                                      $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 8 at 22:37







                                                    • 2




                                                      $begingroup$
                                                      @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
                                                      Mar 9 at 3:23






                                                    • 2




                                                      $begingroup$
                                                      The OP was asking for an argument, not for a "simplest level" that amounts to an unsupported assertion of the same thing he wanted a proof of.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 9 at 10:14














                                                    1












                                                    1








                                                    1





                                                    $begingroup$

                                                    Logically, $y=x$ is a straight line, $y=log_b x$ is not (otherwise why would we call it non-linear function) so they cannot coincide for all $x$.



                                                    Suppose $y=log_b x$ is a straight line the same as $y=x$ then for any two $x_1$ and $x_2$ such as $x_2=x_1+1$ we should have for the slope: $1=fraclog_b x_2-log_b x_1x_2-x_1=log_b x_2-log_b x_1=log_b fracx_2x_1 rightarrow fracx_2x_1=b$ which leads to a contradiction ($frac43=b, frac54=b$).






                                                    share|cite|improve this answer











                                                    $endgroup$



                                                    Logically, $y=x$ is a straight line, $y=log_b x$ is not (otherwise why would we call it non-linear function) so they cannot coincide for all $x$.



                                                    Suppose $y=log_b x$ is a straight line the same as $y=x$ then for any two $x_1$ and $x_2$ such as $x_2=x_1+1$ we should have for the slope: $1=fraclog_b x_2-log_b x_1x_2-x_1=log_b x_2-log_b x_1=log_b fracx_2x_1 rightarrow fracx_2x_1=b$ which leads to a contradiction ($frac43=b, frac54=b$).







                                                    share|cite|improve this answer














                                                    share|cite|improve this answer



                                                    share|cite|improve this answer








                                                    edited Mar 9 at 14:25

























                                                    answered Mar 8 at 22:15









                                                    VasyaVasya

                                                    4,2671618




                                                    4,2671618







                                                    • 2




                                                      $begingroup$
                                                      And why is $y=log_b x$ not a straight line?
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 8 at 22:26






                                                    • 1




                                                      $begingroup$
                                                      @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
                                                      Mar 8 at 22:32










                                                    • $begingroup$
                                                      $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 8 at 22:37







                                                    • 2




                                                      $begingroup$
                                                      @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
                                                      Mar 9 at 3:23






                                                    • 2




                                                      $begingroup$
                                                      The OP was asking for an argument, not for a "simplest level" that amounts to an unsupported assertion of the same thing he wanted a proof of.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 9 at 10:14













                                                    • 2




                                                      $begingroup$
                                                      And why is $y=log_b x$ not a straight line?
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 8 at 22:26






                                                    • 1




                                                      $begingroup$
                                                      @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
                                                      Mar 8 at 22:32










                                                    • $begingroup$
                                                      $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 8 at 22:37







                                                    • 2




                                                      $begingroup$
                                                      @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
                                                      Mar 9 at 3:23






                                                    • 2




                                                      $begingroup$
                                                      The OP was asking for an argument, not for a "simplest level" that amounts to an unsupported assertion of the same thing he wanted a proof of.
                                                      $endgroup$
                                                      – Henning Makholm
                                                      Mar 9 at 10:14








                                                    2




                                                    2




                                                    $begingroup$
                                                    And why is $y=log_b x$ not a straight line?
                                                    $endgroup$
                                                    – Henning Makholm
                                                    Mar 8 at 22:26




                                                    $begingroup$
                                                    And why is $y=log_b x$ not a straight line?
                                                    $endgroup$
                                                    – Henning Makholm
                                                    Mar 8 at 22:26




                                                    1




                                                    1




                                                    $begingroup$
                                                    @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
                                                    Mar 8 at 22:32




                                                    $begingroup$
                                                    @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
                                                    Mar 8 at 22:32












                                                    $begingroup$
                                                    $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                    $endgroup$
                                                    – Henning Makholm
                                                    Mar 8 at 22:37





                                                    $begingroup$
                                                    $frac0x$ is a constant function on a useful subset of $mathbb R$.
                                                    $endgroup$
                                                    – Henning Makholm
                                                    Mar 8 at 22:37





                                                    2




                                                    2




                                                    $begingroup$
                                                    @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
                                                    Mar 9 at 3:23




                                                    $begingroup$
                                                    @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
                                                    Mar 9 at 3:23




                                                    2




                                                    2




                                                    $begingroup$
                                                    The OP was asking for an argument, not for a "simplest level" that amounts to an unsupported assertion of the same thing he wanted a proof of.
                                                    $endgroup$
                                                    – Henning Makholm
                                                    Mar 9 at 10:14





                                                    $begingroup$
                                                    The OP was asking for an argument, not for a "simplest level" that amounts to an unsupported assertion of the same thing he wanted a proof of.
                                                    $endgroup$
                                                    – Henning Makholm
                                                    Mar 9 at 10:14


















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