Solving continuous logarithm whith unknown base and unknown exponent












1












$begingroup$


I was wondering if there is any feasible solution for finding the base and exponent of a continuous logarithm when only the result is known. Thus:



$v=log_{b}e$



with only $v$ given and $b,e$ are integers greater 1.
By feasible I mean an approach other/faster than brute-force.






logarithms







share|cite|improve this question









$endgroup$












  • $begingroup$
    $$qquad e=b^vqquad$$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 31 '18 at 0:00
















1












$begingroup$


I was wondering if there is any feasible solution for finding the base and exponent of a continuous logarithm when only the result is known. Thus:



$v=log_{b}e$



with only $v$ given and $b,e$ are integers greater 1.
By feasible I mean an approach other/faster than brute-force.






logarithms







share|cite|improve this question









$endgroup$












  • $begingroup$
    $$qquad e=b^vqquad$$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 31 '18 at 0:00














1












1








1





$begingroup$


I was wondering if there is any feasible solution for finding the base and exponent of a continuous logarithm when only the result is known. Thus:



$v=log_{b}e$



with only $v$ given and $b,e$ are integers greater 1.
By feasible I mean an approach other/faster than brute-force.






logarithms







share|cite|improve this question









$endgroup$




I was wondering if there is any feasible solution for finding the base and exponent of a continuous logarithm when only the result is known. Thus:



$v=log_{b}e$



with only $v$ given and $b,e$ are integers greater 1.
By feasible I mean an approach other/faster than brute-force.






logarithms




logarithms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 30 '18 at 23:47









AlexAlex

82




82












  • $begingroup$
    $$qquad e=b^vqquad$$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 31 '18 at 0:00


















  • $begingroup$
    $$qquad e=b^vqquad$$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 31 '18 at 0:00
















$begingroup$
$$qquad e=b^vqquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 0:00




$begingroup$
$$qquad e=b^vqquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 0:00










2 Answers
2






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oldest

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4












$begingroup$

Not really, as there are infinite solutions if both $b$ and $e$ are undetermined. Consider, as an example:



$2= log_{10}100$, but also $2= log_{3}9$



In general, $v= log_{b}b^v$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Using natural logarithms (the only ones I know)
    $$v=log_{b}(e) implies v=frac{log (e)}{log (b)}$$ So, if only $v$ is known, you have an infinite number of possible combinations of $(b,e)$.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






      active

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      4












      $begingroup$

      Not really, as there are infinite solutions if both $b$ and $e$ are undetermined. Consider, as an example:



      $2= log_{10}100$, but also $2= log_{3}9$



      In general, $v= log_{b}b^v$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Not really, as there are infinite solutions if both $b$ and $e$ are undetermined. Consider, as an example:



        $2= log_{10}100$, but also $2= log_{3}9$



        In general, $v= log_{b}b^v$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Not really, as there are infinite solutions if both $b$ and $e$ are undetermined. Consider, as an example:



          $2= log_{10}100$, but also $2= log_{3}9$



          In general, $v= log_{b}b^v$






          share|cite|improve this answer









          $endgroup$



          Not really, as there are infinite solutions if both $b$ and $e$ are undetermined. Consider, as an example:



          $2= log_{10}100$, but also $2= log_{3}9$



          In general, $v= log_{b}b^v$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 23:53









          pendermathpendermath

          56812




          56812























              0












              $begingroup$

              Using natural logarithms (the only ones I know)
              $$v=log_{b}(e) implies v=frac{log (e)}{log (b)}$$ So, if only $v$ is known, you have an infinite number of possible combinations of $(b,e)$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Using natural logarithms (the only ones I know)
                $$v=log_{b}(e) implies v=frac{log (e)}{log (b)}$$ So, if only $v$ is known, you have an infinite number of possible combinations of $(b,e)$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Using natural logarithms (the only ones I know)
                  $$v=log_{b}(e) implies v=frac{log (e)}{log (b)}$$ So, if only $v$ is known, you have an infinite number of possible combinations of $(b,e)$.






                  share|cite|improve this answer









                  $endgroup$



                  Using natural logarithms (the only ones I know)
                  $$v=log_{b}(e) implies v=frac{log (e)}{log (b)}$$ So, if only $v$ is known, you have an infinite number of possible combinations of $(b,e)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 31 '18 at 3:47









                  Claude LeiboviciClaude Leibovici

                  125k1158135




                  125k1158135






























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