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$a leq b$, is $a^b leq b^a$ correct?

limit of $x cot x$ as $xto 0$.How to find $dA$ for change of variables.What is $sin(fracpi2-fracnpi2)$ equal to?Finding volume of solid revolution about x-axisCan you take the derivative of a function at infinity?Are points of the form $overlinex = (1,x_2)$ extremes of $f(x) = (x_1-1)^2x_2$?Compute $lim_xto01 - e^-xover e^x - 1$How would I calculate the limit of $g(x)$ that inside another function?graphical meaning of integration by parts of this functionWhy we can't differentiate both sides of a polynomial equation?

4

$begingroup$

I have just thought about an interview question, it was maybe asked previously, but I thought about it myself.

Consider, $a leq b$, is $a^b leq b^a$ correct? Justify.

I thought about solving it in the following manner, but don't end with a conclusive result.

Let's consider:
$$beginalign
a^b &leq b^a \
e^ln(a^b) &leq e^ln(b^a) \
ln(a^b) &leq ln(b^a) \
bln(a) &leq aln(b)
endalign
$$
I know that $ln(a) leq ln(b)$, but cannot conclude from there.
Do you have any suggestion?

calculus

share|cite|improve this question

edited Mar 29 at 21:04

Ertxiem

661112

asked Mar 29 at 20:41

QFiQFi

595315

$endgroup$

add a comment | 

4

$begingroup$

I have just thought about an interview question, it was maybe asked previously, but I thought about it myself.

Consider, $a leq b$, is $a^b leq b^a$ correct? Justify.

I thought about solving it in the following manner, but don't end with a conclusive result.

Let's consider:
$$beginalign
a^b &leq b^a \
e^ln(a^b) &leq e^ln(b^a) \
ln(a^b) &leq ln(b^a) \
bln(a) &leq aln(b)
endalign
$$
I know that $ln(a) leq ln(b)$, but cannot conclude from there.
Do you have any suggestion?

calculus

share|cite|improve this question

edited Mar 29 at 21:04

Ertxiem

661112

asked Mar 29 at 20:41

QFiQFi

595315

$endgroup$

add a comment | 

$begingroup$

I have just thought about an interview question, it was maybe asked previously, but I thought about it myself.

Consider, $a leq b$, is $a^b leq b^a$ correct? Justify.

I thought about solving it in the following manner, but don't end with a conclusive result.

Let's consider:
$$beginalign
a^b &leq b^a \
e^ln(a^b) &leq e^ln(b^a) \
ln(a^b) &leq ln(b^a) \
bln(a) &leq aln(b)
endalign
$$
I know that $ln(a) leq ln(b)$, but cannot conclude from there.
Do you have any suggestion?

calculus

share|cite|improve this question

edited Mar 29 at 21:04

Ertxiem

661112

asked Mar 29 at 20:41

QFiQFi

595315

$endgroup$

share|cite|improve this question

edited Mar 29 at 21:04

Ertxiem

661112

asked Mar 29 at 20:41

QFiQFi

595315

share|cite|improve this question

edited Mar 29 at 21:04

Ertxiem

661112

asked Mar 29 at 20:41

QFiQFi

595315

edited Mar 29 at 21:04

Ertxiem

661112

edited Mar 29 at 21:04

Ertxiem

661112

asked Mar 29 at 20:41

QFiQFi

595315

asked Mar 29 at 20:41

QFiQFi

595315

3 Answers
3

active

oldest

votes

8

$begingroup$

Here is a slightly more systematic way of looking at it.
You want to check whether
$$a le b implies fraclog aa le fraclog bb$$
This would be true if the function $f: mathbbR to mathbbR$
mapping $x mapsto fraclog xx$ were an increasing function.

Now, if you know a little calculus, we can look at the derivative to
get a bit of a clue. It's hopefully easy for you to check that
$$f'(x) = frac1-log xx^2.$$

Now the relationship between increasing/decreasing functions and their
derivatives is the sign of the derivative. It is also hopefully easy
to see that
$$f'(x) text is begincases
textpositive & text if x < e \
textzero & text if x = e \
textnegative & text if x > e
endcases$$

So, $f$ is increasing, stationary, and decreasing on the respective intervals.
Oh no! If $f$ is decreasing, then our desired property will fail. So pick some
$e < a < b$, and you should get a counterexample. Indeed,
$$3 le 4 text but 3^4 > 4^3$$
as (un)desired.

share|cite|improve this answer

answered Mar 29 at 20:57

preferred_anonpreferred_anon

13.1k11843

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This has the nice side effect of proving a fact that looks almost like the original proposition: if $e leq a leq b$ then $b^a leq a^b.$
  $endgroup$
  – David K
  Mar 29 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this crystal clear answer. Indeed, I missed to check the derivative of $f(x)=fraclog(x)x$, as it gives all information needed to have a complete answer to the problem.
  $endgroup$
  – QFi
  Mar 29 at 21:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zizou23 No problem! Glad you found it helpful.
  $endgroup$
  – preferred_anon
  Mar 29 at 23:50
  

  
  
  
  

add a comment | 

3

$begingroup$

Counter-example:

$2le 5$, yet $;2^5>5^2$.

share|cite|improve this answer

edited Mar 29 at 20:48

answered Mar 29 at 20:44

BernardBernard

124k741116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Sorry for the mistyping.
  $endgroup$
  – Bernard
  Mar 29 at 20:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly the problem is not formulated perfectly, as it just popped up in my mind. More opened question would be how do you compare $a^b$ to $b^a$ for $aleq b$. Then, if it is possible to find general solutions for different constraints if required, such as $aleq b leq 0$ or $0leq a leq b$, for instance.
  $endgroup$
  – QFi
  Mar 29 at 20:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer is perfectly valid to the question I posted. Can you edit it in such a way it answers to the following question: $aleq b$ how do you compare $a^b$ to $b^a$? Thinking this question would be more interesting to the community.
  $endgroup$
  – QFi
  Mar 29 at 20:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. However, speaking of $a^b$ and $b^a$ implies $a$ and $b$ are positive if they're not necessarily integers. Also comparing these powers amounts to $fracln a alefracln bb$, i.e. you have to determine whether the function $fracln xx$ is increasing or decreasing. You can do that looking at the derivative.
  $endgroup$
  – Bernard
  Mar 29 at 20:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, thank you.
  $endgroup$
  – QFi
  Mar 29 at 21:08
  

  
  
  
  

add a comment | 

1

$begingroup$

When looking for counterexamples, I recommend trying the simplest cases first.

Are the signs of $a, b$ restricted in any way? If not, try $a = -1, b = 2$.

share|cite|improve this answer

answered Mar 29 at 20:45

avsavs

3,904515

$endgroup$

add a comment | 

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8

$begingroup$

Here is a slightly more systematic way of looking at it.
You want to check whether
$$a le b implies fraclog aa le fraclog bb$$
This would be true if the function $f: mathbbR to mathbbR$
mapping $x mapsto fraclog xx$ were an increasing function.

Now, if you know a little calculus, we can look at the derivative to
get a bit of a clue. It's hopefully easy for you to check that
$$f'(x) = frac1-log xx^2.$$

Now the relationship between increasing/decreasing functions and their
derivatives is the sign of the derivative. It is also hopefully easy
to see that
$$f'(x) text is begincases
textpositive & text if x < e \
textzero & text if x = e \
textnegative & text if x > e
endcases$$

So, $f$ is increasing, stationary, and decreasing on the respective intervals.
Oh no! If $f$ is decreasing, then our desired property will fail. So pick some
$e < a < b$, and you should get a counterexample. Indeed,
$$3 le 4 text but 3^4 > 4^3$$
as (un)desired.

share|cite|improve this answer

answered Mar 29 at 20:57

preferred_anonpreferred_anon

13.1k11843

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This has the nice side effect of proving a fact that looks almost like the original proposition: if $e leq a leq b$ then $b^a leq a^b.$
  $endgroup$
  – David K
  Mar 29 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this crystal clear answer. Indeed, I missed to check the derivative of $f(x)=fraclog(x)x$, as it gives all information needed to have a complete answer to the problem.
  $endgroup$
  – QFi
  Mar 29 at 21:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zizou23 No problem! Glad you found it helpful.
  $endgroup$
  – preferred_anon
  Mar 29 at 23:50
  

  
  
  
  

add a comment | 

8

$begingroup$

Here is a slightly more systematic way of looking at it.
You want to check whether
$$a le b implies fraclog aa le fraclog bb$$
This would be true if the function $f: mathbbR to mathbbR$
mapping $x mapsto fraclog xx$ were an increasing function.

Now, if you know a little calculus, we can look at the derivative to
get a bit of a clue. It's hopefully easy for you to check that
$$f'(x) = frac1-log xx^2.$$

Now the relationship between increasing/decreasing functions and their
derivatives is the sign of the derivative. It is also hopefully easy
to see that
$$f'(x) text is begincases
textpositive & text if x < e \
textzero & text if x = e \
textnegative & text if x > e
endcases$$

So, $f$ is increasing, stationary, and decreasing on the respective intervals.
Oh no! If $f$ is decreasing, then our desired property will fail. So pick some
$e < a < b$, and you should get a counterexample. Indeed,
$$3 le 4 text but 3^4 > 4^3$$
as (un)desired.

share|cite|improve this answer

answered Mar 29 at 20:57

preferred_anonpreferred_anon

13.1k11843

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This has the nice side effect of proving a fact that looks almost like the original proposition: if $e leq a leq b$ then $b^a leq a^b.$
  $endgroup$
  – David K
  Mar 29 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this crystal clear answer. Indeed, I missed to check the derivative of $f(x)=fraclog(x)x$, as it gives all information needed to have a complete answer to the problem.
  $endgroup$
  – QFi
  Mar 29 at 21:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zizou23 No problem! Glad you found it helpful.
  $endgroup$
  – preferred_anon
  Mar 29 at 23:50
  

  
  
  
  

add a comment | 

$begingroup$

Here is a slightly more systematic way of looking at it.
You want to check whether
$$a le b implies fraclog aa le fraclog bb$$
This would be true if the function $f: mathbbR to mathbbR$
mapping $x mapsto fraclog xx$ were an increasing function.

Now, if you know a little calculus, we can look at the derivative to
get a bit of a clue. It's hopefully easy for you to check that
$$f'(x) = frac1-log xx^2.$$

Now the relationship between increasing/decreasing functions and their
derivatives is the sign of the derivative. It is also hopefully easy
to see that
$$f'(x) text is begincases
textpositive & text if x < e \
textzero & text if x = e \
textnegative & text if x > e
endcases$$

So, $f$ is increasing, stationary, and decreasing on the respective intervals.
Oh no! If $f$ is decreasing, then our desired property will fail. So pick some
$e < a < b$, and you should get a counterexample. Indeed,
$$3 le 4 text but 3^4 > 4^3$$
as (un)desired.

share|cite|improve this answer

answered Mar 29 at 20:57

preferred_anonpreferred_anon

13.1k11843

$endgroup$

share|cite|improve this answer

answered Mar 29 at 20:57

preferred_anonpreferred_anon

13.1k11843

answered Mar 29 at 20:57

preferred_anonpreferred_anon

13.1k11843

answered Mar 29 at 20:57

preferred_anonpreferred_anon

13.1k11843

3

$begingroup$

Counter-example:

$2le 5$, yet $;2^5>5^2$.

share|cite|improve this answer

edited Mar 29 at 20:48

answered Mar 29 at 20:44

BernardBernard

124k741116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Sorry for the mistyping.
  $endgroup$
  – Bernard
  Mar 29 at 20:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly the problem is not formulated perfectly, as it just popped up in my mind. More opened question would be how do you compare $a^b$ to $b^a$ for $aleq b$. Then, if it is possible to find general solutions for different constraints if required, such as $aleq b leq 0$ or $0leq a leq b$, for instance.
  $endgroup$
  – QFi
  Mar 29 at 20:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer is perfectly valid to the question I posted. Can you edit it in such a way it answers to the following question: $aleq b$ how do you compare $a^b$ to $b^a$? Thinking this question would be more interesting to the community.
  $endgroup$
  – QFi
  Mar 29 at 20:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. However, speaking of $a^b$ and $b^a$ implies $a$ and $b$ are positive if they're not necessarily integers. Also comparing these powers amounts to $fracln a alefracln bb$, i.e. you have to determine whether the function $fracln xx$ is increasing or decreasing. You can do that looking at the derivative.
  $endgroup$
  – Bernard
  Mar 29 at 20:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, thank you.
  $endgroup$
  – QFi
  Mar 29 at 21:08
  

  
  
  
  

add a comment | 

3

$begingroup$

Counter-example:

$2le 5$, yet $;2^5>5^2$.

share|cite|improve this answer

edited Mar 29 at 20:48

answered Mar 29 at 20:44

BernardBernard

124k741116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Sorry for the mistyping.
  $endgroup$
  – Bernard
  Mar 29 at 20:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly the problem is not formulated perfectly, as it just popped up in my mind. More opened question would be how do you compare $a^b$ to $b^a$ for $aleq b$. Then, if it is possible to find general solutions for different constraints if required, such as $aleq b leq 0$ or $0leq a leq b$, for instance.
  $endgroup$
  – QFi
  Mar 29 at 20:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer is perfectly valid to the question I posted. Can you edit it in such a way it answers to the following question: $aleq b$ how do you compare $a^b$ to $b^a$? Thinking this question would be more interesting to the community.
  $endgroup$
  – QFi
  Mar 29 at 20:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. However, speaking of $a^b$ and $b^a$ implies $a$ and $b$ are positive if they're not necessarily integers. Also comparing these powers amounts to $fracln a alefracln bb$, i.e. you have to determine whether the function $fracln xx$ is increasing or decreasing. You can do that looking at the derivative.
  $endgroup$
  – Bernard
  Mar 29 at 20:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, thank you.
  $endgroup$
  – QFi
  Mar 29 at 21:08
  

  
  
  
  

add a comment | 

$begingroup$

Counter-example:

$2le 5$, yet $;2^5>5^2$.

share|cite|improve this answer

edited Mar 29 at 20:48

answered Mar 29 at 20:44

BernardBernard

124k741116

$endgroup$

share|cite|improve this answer

edited Mar 29 at 20:48

answered Mar 29 at 20:44

BernardBernard

124k741116

answered Mar 29 at 20:44

BernardBernard

124k741116

answered Mar 29 at 20:44

BernardBernard

124k741116

1

$begingroup$

When looking for counterexamples, I recommend trying the simplest cases first.

Are the signs of $a, b$ restricted in any way? If not, try $a = -1, b = 2$.

share|cite|improve this answer

answered Mar 29 at 20:45

avsavs

3,904515

$endgroup$

add a comment | 

1

$begingroup$

When looking for counterexamples, I recommend trying the simplest cases first.

Are the signs of $a, b$ restricted in any way? If not, try $a = -1, b = 2$.

share|cite|improve this answer

answered Mar 29 at 20:45

avsavs

3,904515

$endgroup$

add a comment | 

$begingroup$

When looking for counterexamples, I recommend trying the simplest cases first.

Are the signs of $a, b$ restricted in any way? If not, try $a = -1, b = 2$.

share|cite|improve this answer

answered Mar 29 at 20:45

avsavs

3,904515

$endgroup$

share|cite|improve this answer

answered Mar 29 at 20:45

avsavs

3,904515

answered Mar 29 at 20:45

avsavs

3,904515

answered Mar 29 at 20:45

avsavs

3,904515