Mixing
Comparing powers
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Is there any way to check besides acctually calculating whether $n^x$ is $> = <$ then $m^y$?
For example how to check wheter $440902^532446 > = < 555151^523163$?
exponentiation number-comparison
edited Jul 16 at 22:49
GNU Supporter
11.8k72143
asked Jul 16 at 22:43
Matej Novosad
83
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up vote
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Is there any way to check besides acctually calculating whether $n^x$ is $> = <$ then $m^y$?
For example how to check wheter $440902^532446 > = < 555151^523163$?
exponentiation number-comparison
edited Jul 16 at 22:49
GNU Supporter
11.8k72143
asked Jul 16 at 22:43
Matej Novosad
83
7
Without any restrictions on $m,n,x,y$, I don't think there's any better way other than comparing the log of both sides.
– GNU Supporter
Jul 16 at 22:52
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there any way to check besides acctually calculating whether $n^x$ is $> = <$ then $m^y$?
For example how to check wheter $440902^532446 > = < 555151^523163$?
exponentiation number-comparison
edited Jul 16 at 22:49
GNU Supporter
11.8k72143
asked Jul 16 at 22:43
Matej Novosad
83
Is there any way to check besides acctually calculating whether $n^x$ is $> = <$ then $m^y$?
For example how to check wheter $440902^532446 > = < 555151^523163$?
exponentiation number-comparison
edited Jul 16 at 22:49
GNU Supporter
11.8k72143
asked Jul 16 at 22:43
Matej Novosad
83
edited Jul 16 at 22:49
GNU Supporter
11.8k72143
edited Jul 16 at 22:49
GNU Supporter
11.8k72143
edited Jul 16 at 22:49
GNU Supporter
11.8k72143
11.8k72143
asked Jul 16 at 22:43
Matej Novosad
83
asked Jul 16 at 22:43
Matej Novosad
83
asked Jul 16 at 22:43
Matej Novosad
83
83
7
Without any restrictions on $m,n,x,y$, I don't think there's any better way other than comparing the log of both sides.
– GNU Supporter
Jul 16 at 22:52
add a comment |Â
7
Without any restrictions on $m,n,x,y$, I don't think there's any better way other than comparing the log of both sides.
– GNU Supporter
Jul 16 at 22:52
7
7
Without any restrictions on $m,n,x,y$, I don't think there's any better way other than comparing the log of both sides.
– GNU Supporter
Jul 16 at 22:52
Without any restrictions on $m,n,x,y$, I don't think there's any better way other than comparing the log of both sides.
– GNU Supporter
Jul 16 at 22:52
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
You can make the calculation easier (or at least put it on a more eyeball-able scale) by using logarithms or other monotonic functions. Basically, you want to use a function $f$ such that $x > y$ and $f(x) > f(y)$ are equivalent.
When dealing with exponentiation, logarithms are particularly useful since you can do a lot of manipulations with them. For example:
$log(440902^532446) = 532446log440902$ (which is true regardless of the base of the log)
Or possibly even more useful:
$log_10440902 = log_10100000 + log_104.40902 = 5 + log_104.40902$
So the log, base 10, of the left-hand side of your question is $approx 532446(5 + log_104)$ while the log of the right-hand side is $approx 523163(5 + log_105)$. Since the logs of 4 and 5 are both going to be about 3-ish, the left-hand side has got an obvious lead.
answered Jul 16 at 23:19
ConMan
6,9451324
1
"The logs of $4$ and $5$ are both $3$-ish"? They're $0.602$ and $0.699$, respectively, to three digits. The difference between the actual number of digits between the two offered numbers is only $44$, which is about one part in $12$ thousand of the exponents. Approximating $4.40902$ and $5.55151$ as $4$ and $5$ yields the correct answer, but the error in doing so is about $25$ to $30$ times the actual difference, so the correct answer is attributable mostly to chance.
– Brian Tung
Jul 17 at 0:19
add a comment |Â
up vote
0
down vote
Intuitively, the exponent matters much more than the base, so without computation I would guess the left side is greater. Alpha confirms this. The log of the left is over $100$ more than the log of the right, so the left is greater than the right by a factor over $e^100 approx 2.7cdot 10^43$. To be surer, take the log of each side.
answered Jul 16 at 23:15
Ross Millikan
276k21187352
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You can make the calculation easier (or at least put it on a more eyeball-able scale) by using logarithms or other monotonic functions. Basically, you want to use a function $f$ such that $x > y$ and $f(x) > f(y)$ are equivalent.
When dealing with exponentiation, logarithms are particularly useful since you can do a lot of manipulations with them. For example:
$log(440902^532446) = 532446log440902$ (which is true regardless of the base of the log)
Or possibly even more useful:
$log_10440902 = log_10100000 + log_104.40902 = 5 + log_104.40902$
So the log, base 10, of the left-hand side of your question is $approx 532446(5 + log_104)$ while the log of the right-hand side is $approx 523163(5 + log_105)$. Since the logs of 4 and 5 are both going to be about 3-ish, the left-hand side has got an obvious lead.
answered Jul 16 at 23:19
ConMan
6,9451324
1
"The logs of $4$ and $5$ are both $3$-ish"? They're $0.602$ and $0.699$, respectively, to three digits. The difference between the actual number of digits between the two offered numbers is only $44$, which is about one part in $12$ thousand of the exponents. Approximating $4.40902$ and $5.55151$ as $4$ and $5$ yields the correct answer, but the error in doing so is about $25$ to $30$ times the actual difference, so the correct answer is attributable mostly to chance.
– Brian Tung
Jul 17 at 0:19
add a comment |Â
up vote
0
down vote
accepted
You can make the calculation easier (or at least put it on a more eyeball-able scale) by using logarithms or other monotonic functions. Basically, you want to use a function $f$ such that $x > y$ and $f(x) > f(y)$ are equivalent.
When dealing with exponentiation, logarithms are particularly useful since you can do a lot of manipulations with them. For example:
$log(440902^532446) = 532446log440902$ (which is true regardless of the base of the log)
Or possibly even more useful:
$log_10440902 = log_10100000 + log_104.40902 = 5 + log_104.40902$
So the log, base 10, of the left-hand side of your question is $approx 532446(5 + log_104)$ while the log of the right-hand side is $approx 523163(5 + log_105)$. Since the logs of 4 and 5 are both going to be about 3-ish, the left-hand side has got an obvious lead.
answered Jul 16 at 23:19
ConMan
6,9451324
1
"The logs of $4$ and $5$ are both $3$-ish"? They're $0.602$ and $0.699$, respectively, to three digits. The difference between the actual number of digits between the two offered numbers is only $44$, which is about one part in $12$ thousand of the exponents. Approximating $4.40902$ and $5.55151$ as $4$ and $5$ yields the correct answer, but the error in doing so is about $25$ to $30$ times the actual difference, so the correct answer is attributable mostly to chance.
– Brian Tung
Jul 17 at 0:19
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You can make the calculation easier (or at least put it on a more eyeball-able scale) by using logarithms or other monotonic functions. Basically, you want to use a function $f$ such that $x > y$ and $f(x) > f(y)$ are equivalent.
When dealing with exponentiation, logarithms are particularly useful since you can do a lot of manipulations with them. For example:
$log(440902^532446) = 532446log440902$ (which is true regardless of the base of the log)
Or possibly even more useful:
$log_10440902 = log_10100000 + log_104.40902 = 5 + log_104.40902$
So the log, base 10, of the left-hand side of your question is $approx 532446(5 + log_104)$ while the log of the right-hand side is $approx 523163(5 + log_105)$. Since the logs of 4 and 5 are both going to be about 3-ish, the left-hand side has got an obvious lead.
answered Jul 16 at 23:19
ConMan
6,9451324
You can make the calculation easier (or at least put it on a more eyeball-able scale) by using logarithms or other monotonic functions. Basically, you want to use a function $f$ such that $x > y$ and $f(x) > f(y)$ are equivalent.
When dealing with exponentiation, logarithms are particularly useful since you can do a lot of manipulations with them. For example:
$log(440902^532446) = 532446log440902$ (which is true regardless of the base of the log)
Or possibly even more useful:
$log_10440902 = log_10100000 + log_104.40902 = 5 + log_104.40902$
So the log, base 10, of the left-hand side of your question is $approx 532446(5 + log_104)$ while the log of the right-hand side is $approx 523163(5 + log_105)$. Since the logs of 4 and 5 are both going to be about 3-ish, the left-hand side has got an obvious lead.
answered Jul 16 at 23:19
ConMan
6,9451324
answered Jul 16 at 23:19
ConMan
6,9451324
answered Jul 16 at 23:19
ConMan
6,9451324
answered Jul 16 at 23:19
ConMan
6,9451324
6,9451324
1
"The logs of $4$ and $5$ are both $3$-ish"? They're $0.602$ and $0.699$, respectively, to three digits. The difference between the actual number of digits between the two offered numbers is only $44$, which is about one part in $12$ thousand of the exponents. Approximating $4.40902$ and $5.55151$ as $4$ and $5$ yields the correct answer, but the error in doing so is about $25$ to $30$ times the actual difference, so the correct answer is attributable mostly to chance.
– Brian Tung
Jul 17 at 0:19
add a comment |Â
1
"The logs of $4$ and $5$ are both $3$-ish"? They're $0.602$ and $0.699$, respectively, to three digits. The difference between the actual number of digits between the two offered numbers is only $44$, which is about one part in $12$ thousand of the exponents. Approximating $4.40902$ and $5.55151$ as $4$ and $5$ yields the correct answer, but the error in doing so is about $25$ to $30$ times the actual difference, so the correct answer is attributable mostly to chance.
– Brian Tung
Jul 17 at 0:19
1
1
"The logs of $4$ and $5$ are both $3$-ish"? They're $0.602$ and $0.699$, respectively, to three digits. The difference between the actual number of digits between the two offered numbers is only $44$, which is about one part in $12$ thousand of the exponents. Approximating $4.40902$ and $5.55151$ as $4$ and $5$ yields the correct answer, but the error in doing so is about $25$ to $30$ times the actual difference, so the correct answer is attributable mostly to chance.
– Brian Tung
Jul 17 at 0:19
"The logs of $4$ and $5$ are both $3$-ish"? They're $0.602$ and $0.699$, respectively, to three digits. The difference between the actual number of digits between the two offered numbers is only $44$, which is about one part in $12$ thousand of the exponents. Approximating $4.40902$ and $5.55151$ as $4$ and $5$ yields the correct answer, but the error in doing so is about $25$ to $30$ times the actual difference, so the correct answer is attributable mostly to chance.
– Brian Tung
Jul 17 at 0:19
add a comment |Â
up vote
0
down vote
Intuitively, the exponent matters much more than the base, so without computation I would guess the left side is greater. Alpha confirms this. The log of the left is over $100$ more than the log of the right, so the left is greater than the right by a factor over $e^100 approx 2.7cdot 10^43$. To be surer, take the log of each side.
answered Jul 16 at 23:15
Ross Millikan
276k21187352
add a comment |Â
up vote
0
down vote
Intuitively, the exponent matters much more than the base, so without computation I would guess the left side is greater. Alpha confirms this. The log of the left is over $100$ more than the log of the right, so the left is greater than the right by a factor over $e^100 approx 2.7cdot 10^43$. To be surer, take the log of each side.
answered Jul 16 at 23:15
Ross Millikan
276k21187352
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Intuitively, the exponent matters much more than the base, so without computation I would guess the left side is greater. Alpha confirms this. The log of the left is over $100$ more than the log of the right, so the left is greater than the right by a factor over $e^100 approx 2.7cdot 10^43$. To be surer, take the log of each side.
answered Jul 16 at 23:15
Ross Millikan
276k21187352
Intuitively, the exponent matters much more than the base, so without computation I would guess the left side is greater. Alpha confirms this. The log of the left is over $100$ more than the log of the right, so the left is greater than the right by a factor over $e^100 approx 2.7cdot 10^43$. To be surer, take the log of each side.
answered Jul 16 at 23:15
Ross Millikan
276k21187352
answered Jul 16 at 23:15
Ross Millikan
276k21187352
answered Jul 16 at 23:15
Ross Millikan
276k21187352
answered Jul 16 at 23:15
Ross Millikan
276k21187352
276k21187352
add a comment |Â
add a comment |Â
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7
Without any restrictions on $m,n,x,y$, I don't think there's any better way other than comparing the log of both sides.
– GNU Supporter
Jul 16 at 22:52