Mixing
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Is there any simple ways to compare $x^y$ and $y^x$ without a calculator?
$begingroup$
There are plenty of discussion on MSE about how to compare $x^y$ and $y^x$. For $x,y>e$, it is sufficient to just compare $x$ and $y$ to reach a conclusion. But I wonder if there are some general steps that can be used in any situations where $x<e$ and $y>e$. I tried to use logarithm, but the inequities produced often includes multiplication of negative numbers, so I often make mistakes along the way.
Note that I am interested in all REAL values of $x,y$, NOT just integers. For example, $x=2,y=sqrt{5}$. I wish anyone to give a solution that is applicable in other values of $x,y$ as well.
algebra-precalculus inequality exponentiation number-comparison
asked Feb 2 at 14:55
Holding ArthurHolding Arthur
1,575417
$endgroup$
add a comment |
$begingroup$
There are plenty of discussion on MSE about how to compare $x^y$ and $y^x$. For $x,y>e$, it is sufficient to just compare $x$ and $y$ to reach a conclusion. But I wonder if there are some general steps that can be used in any situations where $x<e$ and $y>e$. I tried to use logarithm, but the inequities produced often includes multiplication of negative numbers, so I often make mistakes along the way.
Note that I am interested in all REAL values of $x,y$, NOT just integers. For example, $x=2,y=sqrt{5}$. I wish anyone to give a solution that is applicable in other values of $x,y$ as well.
algebra-precalculus inequality exponentiation number-comparison
asked Feb 2 at 14:55
Holding ArthurHolding Arthur
1,575417
$endgroup$
$begingroup$
"... all real values of $x,y$ ..." Presumably just positive? Since $x^y$ might be undefined otherwise. Cool question otherwise (+1)
$endgroup$
– MisterRiemann
Feb 2 at 15:43
$begingroup$
Nice question, have a look at this: mathforum.org/library/drmath/view/70271.html
$endgroup$
– anas pcpro
Feb 2 at 17:03
$begingroup$
@MisterRiemann Yes you are right
$endgroup$
– Holding Arthur
Feb 2 at 22:52
$begingroup$
@anaspcpro That link just discusses the case where x, y are integers. It have not considered the case where x, y are generally any real number.
$endgroup$
– Holding Arthur
Feb 2 at 23:20
$begingroup$
That's why I didn't add the link as an answer.
$endgroup$
– anas pcpro
Feb 3 at 16:00
add a comment |
$begingroup$
There are plenty of discussion on MSE about how to compare $x^y$ and $y^x$. For $x,y>e$, it is sufficient to just compare $x$ and $y$ to reach a conclusion. But I wonder if there are some general steps that can be used in any situations where $x<e$ and $y>e$. I tried to use logarithm, but the inequities produced often includes multiplication of negative numbers, so I often make mistakes along the way.
Note that I am interested in all REAL values of $x,y$, NOT just integers. For example, $x=2,y=sqrt{5}$. I wish anyone to give a solution that is applicable in other values of $x,y$ as well.
algebra-precalculus inequality exponentiation number-comparison
asked Feb 2 at 14:55
Holding ArthurHolding Arthur
1,575417
$endgroup$
There are plenty of discussion on MSE about how to compare $x^y$ and $y^x$. For $x,y>e$, it is sufficient to just compare $x$ and $y$ to reach a conclusion. But I wonder if there are some general steps that can be used in any situations where $x<e$ and $y>e$. I tried to use logarithm, but the inequities produced often includes multiplication of negative numbers, so I often make mistakes along the way.
Note that I am interested in all REAL values of $x,y$, NOT just integers. For example, $x=2,y=sqrt{5}$. I wish anyone to give a solution that is applicable in other values of $x,y$ as well.
algebra-precalculus inequality exponentiation number-comparison
algebra-precalculus inequality exponentiation number-comparison
asked Feb 2 at 14:55
Holding ArthurHolding Arthur
1,575417
asked Feb 2 at 14:55
Holding ArthurHolding Arthur
1,575417
asked Feb 2 at 14:55
Holding ArthurHolding Arthur
1,575417
asked Feb 2 at 14:55
Holding ArthurHolding Arthur
1,575417
asked Feb 2 at 14:55
Holding ArthurHolding Arthur
1,575417
1,575417
$begingroup$
"... all real values of $x,y$ ..." Presumably just positive? Since $x^y$ might be undefined otherwise. Cool question otherwise (+1)
$endgroup$
– MisterRiemann
Feb 2 at 15:43
$begingroup$
Nice question, have a look at this: mathforum.org/library/drmath/view/70271.html
$endgroup$
– anas pcpro
Feb 2 at 17:03
$begingroup$
@MisterRiemann Yes you are right
$endgroup$
– Holding Arthur
Feb 2 at 22:52
$begingroup$
@anaspcpro That link just discusses the case where x, y are integers. It have not considered the case where x, y are generally any real number.
$endgroup$
– Holding Arthur
Feb 2 at 23:20
$begingroup$
That's why I didn't add the link as an answer.
$endgroup$
– anas pcpro
Feb 3 at 16:00
add a comment |
$begingroup$
"... all real values of $x,y$ ..." Presumably just positive? Since $x^y$ might be undefined otherwise. Cool question otherwise (+1)
$endgroup$
– MisterRiemann
Feb 2 at 15:43
$begingroup$
Nice question, have a look at this: mathforum.org/library/drmath/view/70271.html
$endgroup$
– anas pcpro
Feb 2 at 17:03
$begingroup$
@MisterRiemann Yes you are right
$endgroup$
– Holding Arthur
Feb 2 at 22:52
$begingroup$
@anaspcpro That link just discusses the case where x, y are integers. It have not considered the case where x, y are generally any real number.
$endgroup$
– Holding Arthur
Feb 2 at 23:20
$begingroup$
That's why I didn't add the link as an answer.
$endgroup$
– anas pcpro
Feb 3 at 16:00
$begingroup$
"... all real values of $x,y$ ..." Presumably just positive? Since $x^y$ might be undefined otherwise. Cool question otherwise (+1)
$endgroup$
– MisterRiemann
Feb 2 at 15:43
$begingroup$
"... all real values of $x,y$ ..." Presumably just positive? Since $x^y$ might be undefined otherwise. Cool question otherwise (+1)
$endgroup$
– MisterRiemann
Feb 2 at 15:43
$begingroup$
Nice question, have a look at this: mathforum.org/library/drmath/view/70271.html
$endgroup$
– anas pcpro
Feb 2 at 17:03
$begingroup$
Nice question, have a look at this: mathforum.org/library/drmath/view/70271.html
$endgroup$
– anas pcpro
Feb 2 at 17:03
$begingroup$
@MisterRiemann Yes you are right
$endgroup$
– Holding Arthur
Feb 2 at 22:52
$begingroup$
@MisterRiemann Yes you are right
$endgroup$
– Holding Arthur
Feb 2 at 22:52
$begingroup$
@anaspcpro That link just discusses the case where x, y are integers. It have not considered the case where x, y are generally any real number.
$endgroup$
– Holding Arthur
Feb 2 at 23:20
$begingroup$
@anaspcpro That link just discusses the case where x, y are integers. It have not considered the case where x, y are generally any real number.
$endgroup$
– Holding Arthur
Feb 2 at 23:20
$begingroup$
That's why I didn't add the link as an answer.
$endgroup$
– anas pcpro
Feb 3 at 16:00
$begingroup$
That's why I didn't add the link as an answer.
$endgroup$
– anas pcpro
Feb 3 at 16:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I can offer a "comparison test".
W.l.o.g., suppose you are interested to establish whether or not $x^y > y^x$. Now set $a=y/x$. Then it is easy to show that $x^y > y^x$ is equivalent to
$$
x > f(a) = a^{frac{1}{a-1}}
$$
Now you can exploit the fact that $f(a)$ is strictly decreasing with $a$. That allows to come up with the following comparison test:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} > y_0^{x_0}$ holds, then $x_0^{y} > y^{x_0}$ will also hold for all $y > y_0$.
And:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} < y_0^{x_0}$ holds, then $x_0^{y} < y^{x_0}$ will also hold for all $y < y_0$.
A particular "test case" is $x_0=y_0$ which gives $f(1) = e$. Then the above statements are:
If $y>x>e$, then $x^{y} > y^{x}$.
And: If $x<y<e$, then $x^{y} < y^{x}$.
answered Mar 4 at 13:55
AndreasAndreas
8,4511137
$endgroup$
$begingroup$
But how can I get $y_0$ at the first place?
$endgroup$
– Holding Arthur
Mar 4 at 14:04
$begingroup$
Well, since we have two special cases where the comparison is easy, cases remain to be investigated only where one of the two variables is less than $e$ and the other is higher than $e$. For these cases, you need initial values to compare. A good start would be to find all value pairs $(x_0,y_0)$ where $x_0^{y_0} = y_0^{x_0}$. Take as an example $(x_0,y_0) = (2,4)$. Then we know that $ 2^{y} > y^2$ for $y >4$. Reversing the pair, we have that $ 4^{y} > y^4$ for $y <2$.
$endgroup$
– Andreas
Mar 5 at 10:24
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I can offer a "comparison test".
W.l.o.g., suppose you are interested to establish whether or not $x^y > y^x$. Now set $a=y/x$. Then it is easy to show that $x^y > y^x$ is equivalent to
$$
x > f(a) = a^{frac{1}{a-1}}
$$
Now you can exploit the fact that $f(a)$ is strictly decreasing with $a$. That allows to come up with the following comparison test:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} > y_0^{x_0}$ holds, then $x_0^{y} > y^{x_0}$ will also hold for all $y > y_0$.
And:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} < y_0^{x_0}$ holds, then $x_0^{y} < y^{x_0}$ will also hold for all $y < y_0$.
A particular "test case" is $x_0=y_0$ which gives $f(1) = e$. Then the above statements are:
If $y>x>e$, then $x^{y} > y^{x}$.
And: If $x<y<e$, then $x^{y} < y^{x}$.
answered Mar 4 at 13:55
AndreasAndreas
8,4511137
$endgroup$
$begingroup$
But how can I get $y_0$ at the first place?
$endgroup$
– Holding Arthur
Mar 4 at 14:04
$begingroup$
Well, since we have two special cases where the comparison is easy, cases remain to be investigated only where one of the two variables is less than $e$ and the other is higher than $e$. For these cases, you need initial values to compare. A good start would be to find all value pairs $(x_0,y_0)$ where $x_0^{y_0} = y_0^{x_0}$. Take as an example $(x_0,y_0) = (2,4)$. Then we know that $ 2^{y} > y^2$ for $y >4$. Reversing the pair, we have that $ 4^{y} > y^4$ for $y <2$.
$endgroup$
– Andreas
Mar 5 at 10:24
add a comment |
$begingroup$
I can offer a "comparison test".
W.l.o.g., suppose you are interested to establish whether or not $x^y > y^x$. Now set $a=y/x$. Then it is easy to show that $x^y > y^x$ is equivalent to
$$
x > f(a) = a^{frac{1}{a-1}}
$$
Now you can exploit the fact that $f(a)$ is strictly decreasing with $a$. That allows to come up with the following comparison test:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} > y_0^{x_0}$ holds, then $x_0^{y} > y^{x_0}$ will also hold for all $y > y_0$.
And:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} < y_0^{x_0}$ holds, then $x_0^{y} < y^{x_0}$ will also hold for all $y < y_0$.
A particular "test case" is $x_0=y_0$ which gives $f(1) = e$. Then the above statements are:
If $y>x>e$, then $x^{y} > y^{x}$.
And: If $x<y<e$, then $x^{y} < y^{x}$.
answered Mar 4 at 13:55
AndreasAndreas
8,4511137
$endgroup$
$begingroup$
But how can I get $y_0$ at the first place?
$endgroup$
– Holding Arthur
Mar 4 at 14:04
$begingroup$
Well, since we have two special cases where the comparison is easy, cases remain to be investigated only where one of the two variables is less than $e$ and the other is higher than $e$. For these cases, you need initial values to compare. A good start would be to find all value pairs $(x_0,y_0)$ where $x_0^{y_0} = y_0^{x_0}$. Take as an example $(x_0,y_0) = (2,4)$. Then we know that $ 2^{y} > y^2$ for $y >4$. Reversing the pair, we have that $ 4^{y} > y^4$ for $y <2$.
$endgroup$
– Andreas
Mar 5 at 10:24
add a comment |
$begingroup$
I can offer a "comparison test".
W.l.o.g., suppose you are interested to establish whether or not $x^y > y^x$. Now set $a=y/x$. Then it is easy to show that $x^y > y^x$ is equivalent to
$$
x > f(a) = a^{frac{1}{a-1}}
$$
Now you can exploit the fact that $f(a)$ is strictly decreasing with $a$. That allows to come up with the following comparison test:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} > y_0^{x_0}$ holds, then $x_0^{y} > y^{x_0}$ will also hold for all $y > y_0$.
And:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} < y_0^{x_0}$ holds, then $x_0^{y} < y^{x_0}$ will also hold for all $y < y_0$.
A particular "test case" is $x_0=y_0$ which gives $f(1) = e$. Then the above statements are:
If $y>x>e$, then $x^{y} > y^{x}$.
And: If $x<y<e$, then $x^{y} < y^{x}$.
answered Mar 4 at 13:55
AndreasAndreas
8,4511137
$endgroup$
I can offer a "comparison test".
W.l.o.g., suppose you are interested to establish whether or not $x^y > y^x$. Now set $a=y/x$. Then it is easy to show that $x^y > y^x$ is equivalent to
$$
x > f(a) = a^{frac{1}{a-1}}
$$
Now you can exploit the fact that $f(a)$ is strictly decreasing with $a$. That allows to come up with the following comparison test:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} > y_0^{x_0}$ holds, then $x_0^{y} > y^{x_0}$ will also hold for all $y > y_0$.
And:
If you have a pair $(x_0,y_0)$ for which $x_0^{y_0} < y_0^{x_0}$ holds, then $x_0^{y} < y^{x_0}$ will also hold for all $y < y_0$.
A particular "test case" is $x_0=y_0$ which gives $f(1) = e$. Then the above statements are:
If $y>x>e$, then $x^{y} > y^{x}$.
And: If $x<y<e$, then $x^{y} < y^{x}$.
answered Mar 4 at 13:55
AndreasAndreas
8,4511137
answered Mar 4 at 13:55
AndreasAndreas
8,4511137
answered Mar 4 at 13:55
AndreasAndreas
8,4511137
answered Mar 4 at 13:55
AndreasAndreas
8,4511137
8,4511137
$begingroup$
But how can I get $y_0$ at the first place?
$endgroup$
– Holding Arthur
Mar 4 at 14:04
$begingroup$
Well, since we have two special cases where the comparison is easy, cases remain to be investigated only where one of the two variables is less than $e$ and the other is higher than $e$. For these cases, you need initial values to compare. A good start would be to find all value pairs $(x_0,y_0)$ where $x_0^{y_0} = y_0^{x_0}$. Take as an example $(x_0,y_0) = (2,4)$. Then we know that $ 2^{y} > y^2$ for $y >4$. Reversing the pair, we have that $ 4^{y} > y^4$ for $y <2$.
$endgroup$
– Andreas
Mar 5 at 10:24
add a comment |
$begingroup$
But how can I get $y_0$ at the first place?
$endgroup$
– Holding Arthur
Mar 4 at 14:04
$begingroup$
Well, since we have two special cases where the comparison is easy, cases remain to be investigated only where one of the two variables is less than $e$ and the other is higher than $e$. For these cases, you need initial values to compare. A good start would be to find all value pairs $(x_0,y_0)$ where $x_0^{y_0} = y_0^{x_0}$. Take as an example $(x_0,y_0) = (2,4)$. Then we know that $ 2^{y} > y^2$ for $y >4$. Reversing the pair, we have that $ 4^{y} > y^4$ for $y <2$.
$endgroup$
– Andreas
Mar 5 at 10:24
$begingroup$
But how can I get $y_0$ at the first place?
$endgroup$
– Holding Arthur
Mar 4 at 14:04
$begingroup$
But how can I get $y_0$ at the first place?
$endgroup$
– Holding Arthur
Mar 4 at 14:04
$begingroup$
Well, since we have two special cases where the comparison is easy, cases remain to be investigated only where one of the two variables is less than $e$ and the other is higher than $e$. For these cases, you need initial values to compare. A good start would be to find all value pairs $(x_0,y_0)$ where $x_0^{y_0} = y_0^{x_0}$. Take as an example $(x_0,y_0) = (2,4)$. Then we know that $ 2^{y} > y^2$ for $y >4$. Reversing the pair, we have that $ 4^{y} > y^4$ for $y <2$.
$endgroup$
– Andreas
Mar 5 at 10:24
$begingroup$
Well, since we have two special cases where the comparison is easy, cases remain to be investigated only where one of the two variables is less than $e$ and the other is higher than $e$. For these cases, you need initial values to compare. A good start would be to find all value pairs $(x_0,y_0)$ where $x_0^{y_0} = y_0^{x_0}$. Take as an example $(x_0,y_0) = (2,4)$. Then we know that $ 2^{y} > y^2$ for $y >4$. Reversing the pair, we have that $ 4^{y} > y^4$ for $y <2$.
$endgroup$
– Andreas
Mar 5 at 10:24
add a comment |
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$begingroup$
"... all real values of $x,y$ ..." Presumably just positive? Since $x^y$ might be undefined otherwise. Cool question otherwise (+1)
$endgroup$
– MisterRiemann
Feb 2 at 15:43
$begingroup$
Nice question, have a look at this: mathforum.org/library/drmath/view/70271.html
$endgroup$
– anas pcpro
Feb 2 at 17:03
$begingroup$
@MisterRiemann Yes you are right
$endgroup$
– Holding Arthur
Feb 2 at 22:52
$begingroup$
@anaspcpro That link just discusses the case where x, y are integers. It have not considered the case where x, y are generally any real number.
$endgroup$
– Holding Arthur
Feb 2 at 23:20
$begingroup$
That's why I didn't add the link as an answer.
$endgroup$
– anas pcpro
Feb 3 at 16:00