Solving continuous logarithm whith unknown base and unknown exponent
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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
asked Dec 30 '18 at 23:47
AlexAlex
82
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add a comment |
$begingroup$
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
asked Dec 30 '18 at 23:47
AlexAlex
82
$endgroup$
$begingroup$
$$qquad e=b^vqquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 0:00
add a comment |
$begingroup$
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
asked Dec 30 '18 at 23:47
AlexAlex
82
$endgroup$
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
asked Dec 30 '18 at 23:47
AlexAlex
82
asked Dec 30 '18 at 23:47
AlexAlex
82
asked Dec 30 '18 at 23:47
AlexAlex
82
asked Dec 30 '18 at 23:47
AlexAlex
82
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
add a comment |
$begingroup$
$$qquad e=b^vqquad$$
$endgroup$
– 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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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, $v= log_{b}b^v$
answered Dec 30 '18 at 23:53
pendermathpendermath
56812
$endgroup$
add a comment |
$begingroup$
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 $(b,e)$.
answered Dec 31 '18 at 3:47
Claude LeiboviciClaude Leibovici
125k1158135
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
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, $v= log_{b}b^v$
answered Dec 30 '18 at 23:53
pendermathpendermath
56812
$endgroup$
add a comment |
$begingroup$
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, $v= log_{b}b^v$
answered Dec 30 '18 at 23:53
pendermathpendermath
56812
$endgroup$
add a comment |
$begingroup$
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, $v= log_{b}b^v$
answered Dec 30 '18 at 23:53
pendermathpendermath
56812
$endgroup$
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, $v= log_{b}b^v$
answered Dec 30 '18 at 23:53
pendermathpendermath
56812
answered Dec 30 '18 at 23:53
pendermathpendermath
56812
answered Dec 30 '18 at 23:53
pendermathpendermath
56812
answered Dec 30 '18 at 23:53
pendermathpendermath
56812
56812
add a comment |
add a comment |
$begingroup$
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 $(b,e)$.
answered Dec 31 '18 at 3:47
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
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 $(b,e)$.
answered Dec 31 '18 at 3:47
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
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 $(b,e)$.
answered Dec 31 '18 at 3:47
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
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 $(b,e)$.
answered Dec 31 '18 at 3:47
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 31 '18 at 3:47
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 31 '18 at 3:47
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 31 '18 at 3:47
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
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$begingroup$
$$qquad e=b^vqquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 0:00