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












1












$begingroup$


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.)





Answer



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.










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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
    2 mins ago


















1












$begingroup$


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.)





Answer



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 question









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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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  • $begingroup$
    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
    2 mins ago
















1












1








1





$begingroup$


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.)





Answer



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 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.)





Answer



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.







logarithms






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edited 1 hour ago







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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
    2 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
    2 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
2 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
2 mins ago












6 Answers
6






active

oldest

votes


















7












$begingroup$

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












        $begingroup$

        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
          4 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
          4 hours ago










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





















        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











          $endgroup$













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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7












            $begingroup$

            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









            $endgroup$


















              7












              $begingroup$

              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









              $endgroup$
















                7












                7








                7





                $begingroup$

                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

                1,248612




                1,248612























                    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












                                    $begingroup$

                                    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
                                      4 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
                                      4 hours ago










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


















                                    1












                                    $begingroup$

                                    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
                                      4 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
                                      4 hours ago










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
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    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 4 hours ago









                                    VasyaVasya

                                    3,9551618




                                    3,9551618








                                    • 1




                                      $begingroup$
                                      And why is $y=log_b x$ not a straight line?
                                      $endgroup$
                                      – Henning Makholm
                                      4 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
                                      4 hours ago










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
















                                    • 1




                                      $begingroup$
                                      And why is $y=log_b x$ not a straight line?
                                      $endgroup$
                                      – Henning Makholm
                                      4 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
                                      4 hours ago










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










                                    1




                                    1




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




                                    $begingroup$
                                    And why is $y=log_b x$ not a straight line?
                                    $endgroup$
                                    – Henning Makholm
                                    4 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
                                    4 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
                                    4 hours ago












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






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













                                    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 3 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 3 hours ago

























                                                answered 3 hours ago









                                                enedilenedil

                                                1,359620




                                                1,359620






















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