Show that $x^a le x$ for $x in (0,1)$ and $a ge 1$

0

$begingroup$

Using the definition $$x^a = e^{a log x} $$ is there any way to show that for $x in (0,1)$ and $a ge 1$ that $x^a le x$.

Using the traditional definition of $x^a$ ($x^a$ = $x$ multiplied by itself $a$ times) the result is obvious. I was wondering if there was any simple way to show this result using the above definition. Any hints would be welcome.

real-analysis

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asked Jan 20 at 2:20

Is12PrimeIs12Prime

135111

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $x = e^{log x}$.
  $endgroup$
  – youngsmasher
  Jan 20 at 2:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is equivalent to showing $x^{a-1}leq 1$
  $endgroup$
  – rtybase
  Jan 20 at 2:26
  

  
  
  
  

add a comment | 

0

$begingroup$

Using the definition $$x^a = e^{a log x} $$ is there any way to show that for $x in (0,1)$ and $a ge 1$ that $x^a le x$.

Using the traditional definition of $x^a$ ($x^a$ = $x$ multiplied by itself $a$ times) the result is obvious. I was wondering if there was any simple way to show this result using the above definition. Any hints would be welcome.

real-analysis

share|cite|improve this question

asked Jan 20 at 2:20

Is12PrimeIs12Prime

135111

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $x = e^{log x}$.
  $endgroup$
  – youngsmasher
  Jan 20 at 2:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is equivalent to showing $x^{a-1}leq 1$
  $endgroup$
  – rtybase
  Jan 20 at 2:26
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

Using the definition $$x^a = e^{a log x} $$ is there any way to show that for $x in (0,1)$ and $a ge 1$ that $x^a le x$.

Using the traditional definition of $x^a$ ($x^a$ = $x$ multiplied by itself $a$ times) the result is obvious. I was wondering if there was any simple way to show this result using the above definition. Any hints would be welcome.

real-analysis

share|cite|improve this question

asked Jan 20 at 2:20

Is12PrimeIs12Prime

135111

$endgroup$

Using the definition $$x^a = e^{a log x} $$ is there any way to show that for $x in (0,1)$ and $a ge 1$ that $x^a le x$.

Using the traditional definition of $x^a$ ($x^a$ = $x$ multiplied by itself $a$ times) the result is obvious. I was wondering if there was any simple way to show this result using the above definition. Any hints would be welcome.

real-analysis

real-analysis

share|cite|improve this question

asked Jan 20 at 2:20

Is12PrimeIs12Prime

135111

share|cite|improve this question

asked Jan 20 at 2:20

Is12PrimeIs12Prime

135111

share|cite|improve this question

share|cite|improve this question

asked Jan 20 at 2:20

Is12PrimeIs12Prime

135111

asked Jan 20 at 2:20

Is12PrimeIs12Prime

135111

asked Jan 20 at 2:20

Is12PrimeIs12Prime

135111

135111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $x = e^{log x}$.
  $endgroup$
  – youngsmasher
  Jan 20 at 2:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is equivalent to showing $x^{a-1}leq 1$
  $endgroup$
  – rtybase
  Jan 20 at 2:26
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $x = e^{log x}$.
  $endgroup$
  – youngsmasher
  Jan 20 at 2:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is equivalent to showing $x^{a-1}leq 1$
  $endgroup$
  – rtybase
  Jan 20 at 2:26
  

  
  
  
  

1

1

$begingroup$
Hint: $x = e^{log x}$.
$endgroup$
– youngsmasher
Jan 20 at 2:23

$begingroup$
Hint: $x = e^{log x}$.
$endgroup$
– youngsmasher
Jan 20 at 2:23

2

2

$begingroup$
This is equivalent to showing $x^{a-1}leq 1$
$endgroup$
– rtybase
Jan 20 at 2:26

$begingroup$
This is equivalent to showing $x^{a-1}leq 1$
$endgroup$
– rtybase
Jan 20 at 2:26

add a comment | 

1 Answer
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2

$begingroup$

Using the requested definition of

$$x^a = e^{alog x} tag{1}label{eq1}$$

note that

$$x^a - x = e^{alog x} - e^{log x} = e^{log x}left(e^{left(a - 1right)log x} - 1right) tag{2}label{eq2}$$

Now,

$$e^{log x} = x gt 0 tag{3}label{eq3}$$

Also, it's given that

$$a ge 1 Rightarrow a - 1 ge 0 tag{4}label{eq4}$$

In addition, note that

$$log x lt 0 text{ for } 0 lt x lt 1 tag{5}label{eq5}$$

Thus, using eqref{eq4}, we get

$$left(a - 1right)log x leq 0 tag{6}label{eq6}$$

giving that

$$e^{left(a - 1right)log x} leq 1 Rightarrow e^{left(a - 1right)log x} - 1 leq 0 tag{7}label{eq7}$$

Putting this and eqref{eq3} into eqref{eq2}, using that you have a positive value multiplied by a value which is $0$ or negative, gives that

$$x^a - x leq 0 Rightarrow x^a leq x tag{8}label{eq8}$$

I hope I haven't used any properties I'm not supposed to.

Nothing above is specific to using a base of $e$ as the statements just use a basic property of logarithms that $log x lt 0$ if $x lt 1$, and that the base to any negative power is $lt 1$, so any other positive base, e.g., $10$, could have been used instead.

share|cite|improve this answer

edited Jan 20 at 2:59

answered Jan 20 at 2:39

John OmielanJohn Omielan

3,4801215

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add a comment | 

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1 Answer
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1 Answer
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2

$begingroup$

Using the requested definition of

$$x^a = e^{alog x} tag{1}label{eq1}$$

note that

$$x^a - x = e^{alog x} - e^{log x} = e^{log x}left(e^{left(a - 1right)log x} - 1right) tag{2}label{eq2}$$

Now,

$$e^{log x} = x gt 0 tag{3}label{eq3}$$

Also, it's given that

$$a ge 1 Rightarrow a - 1 ge 0 tag{4}label{eq4}$$

In addition, note that

$$log x lt 0 text{ for } 0 lt x lt 1 tag{5}label{eq5}$$

Thus, using eqref{eq4}, we get

$$left(a - 1right)log x leq 0 tag{6}label{eq6}$$

giving that

$$e^{left(a - 1right)log x} leq 1 Rightarrow e^{left(a - 1right)log x} - 1 leq 0 tag{7}label{eq7}$$

Putting this and eqref{eq3} into eqref{eq2}, using that you have a positive value multiplied by a value which is $0$ or negative, gives that

$$x^a - x leq 0 Rightarrow x^a leq x tag{8}label{eq8}$$

I hope I haven't used any properties I'm not supposed to.

Nothing above is specific to using a base of $e$ as the statements just use a basic property of logarithms that $log x lt 0$ if $x lt 1$, and that the base to any negative power is $lt 1$, so any other positive base, e.g., $10$, could have been used instead.

share|cite|improve this answer

edited Jan 20 at 2:59

answered Jan 20 at 2:39

John OmielanJohn Omielan

3,4801215

$endgroup$

add a comment | 

2

$begingroup$

Using the requested definition of

$$x^a = e^{alog x} tag{1}label{eq1}$$

note that

$$x^a - x = e^{alog x} - e^{log x} = e^{log x}left(e^{left(a - 1right)log x} - 1right) tag{2}label{eq2}$$

Now,

$$e^{log x} = x gt 0 tag{3}label{eq3}$$

Also, it's given that

$$a ge 1 Rightarrow a - 1 ge 0 tag{4}label{eq4}$$

In addition, note that

$$log x lt 0 text{ for } 0 lt x lt 1 tag{5}label{eq5}$$

Thus, using eqref{eq4}, we get

$$left(a - 1right)log x leq 0 tag{6}label{eq6}$$

giving that

$$e^{left(a - 1right)log x} leq 1 Rightarrow e^{left(a - 1right)log x} - 1 leq 0 tag{7}label{eq7}$$

Putting this and eqref{eq3} into eqref{eq2}, using that you have a positive value multiplied by a value which is $0$ or negative, gives that

$$x^a - x leq 0 Rightarrow x^a leq x tag{8}label{eq8}$$

I hope I haven't used any properties I'm not supposed to.

Nothing above is specific to using a base of $e$ as the statements just use a basic property of logarithms that $log x lt 0$ if $x lt 1$, and that the base to any negative power is $lt 1$, so any other positive base, e.g., $10$, could have been used instead.

share|cite|improve this answer

edited Jan 20 at 2:59

answered Jan 20 at 2:39

John OmielanJohn Omielan

3,4801215

$endgroup$

add a comment | 

2

2

2

$begingroup$

Using the requested definition of

$$x^a = e^{alog x} tag{1}label{eq1}$$

note that

$$x^a - x = e^{alog x} - e^{log x} = e^{log x}left(e^{left(a - 1right)log x} - 1right) tag{2}label{eq2}$$

Now,

$$e^{log x} = x gt 0 tag{3}label{eq3}$$

Also, it's given that

$$a ge 1 Rightarrow a - 1 ge 0 tag{4}label{eq4}$$

In addition, note that

$$log x lt 0 text{ for } 0 lt x lt 1 tag{5}label{eq5}$$

Thus, using eqref{eq4}, we get

$$left(a - 1right)log x leq 0 tag{6}label{eq6}$$

giving that

$$e^{left(a - 1right)log x} leq 1 Rightarrow e^{left(a - 1right)log x} - 1 leq 0 tag{7}label{eq7}$$

Putting this and eqref{eq3} into eqref{eq2}, using that you have a positive value multiplied by a value which is $0$ or negative, gives that

$$x^a - x leq 0 Rightarrow x^a leq x tag{8}label{eq8}$$

I hope I haven't used any properties I'm not supposed to.

Nothing above is specific to using a base of $e$ as the statements just use a basic property of logarithms that $log x lt 0$ if $x lt 1$, and that the base to any negative power is $lt 1$, so any other positive base, e.g., $10$, could have been used instead.

share|cite|improve this answer

edited Jan 20 at 2:59

answered Jan 20 at 2:39

John OmielanJohn Omielan

3,4801215

$endgroup$

Using the requested definition of

$$x^a = e^{alog x} tag{1}label{eq1}$$

note that

$$x^a - x = e^{alog x} - e^{log x} = e^{log x}left(e^{left(a - 1right)log x} - 1right) tag{2}label{eq2}$$

Now,

$$e^{log x} = x gt 0 tag{3}label{eq3}$$

Also, it's given that

$$a ge 1 Rightarrow a - 1 ge 0 tag{4}label{eq4}$$

In addition, note that

$$log x lt 0 text{ for } 0 lt x lt 1 tag{5}label{eq5}$$

Thus, using eqref{eq4}, we get

$$left(a - 1right)log x leq 0 tag{6}label{eq6}$$

giving that

$$e^{left(a - 1right)log x} leq 1 Rightarrow e^{left(a - 1right)log x} - 1 leq 0 tag{7}label{eq7}$$

Putting this and eqref{eq3} into eqref{eq2}, using that you have a positive value multiplied by a value which is $0$ or negative, gives that

$$x^a - x leq 0 Rightarrow x^a leq x tag{8}label{eq8}$$

I hope I haven't used any properties I'm not supposed to.

Nothing above is specific to using a base of $e$ as the statements just use a basic property of logarithms that $log x lt 0$ if $x lt 1$, and that the base to any negative power is $lt 1$, so any other positive base, e.g., $10$, could have been used instead.

share|cite|improve this answer

edited Jan 20 at 2:59

answered Jan 20 at 2:39

John OmielanJohn Omielan

3,4801215

share|cite|improve this answer

share|cite|improve this answer

edited Jan 20 at 2:59

edited Jan 20 at 2:59

edited Jan 20 at 2:59

answered Jan 20 at 2:39

John OmielanJohn Omielan

3,4801215

answered Jan 20 at 2:39

John OmielanJohn Omielan

3,4801215

answered Jan 20 at 2:39

John OmielanJohn Omielan

3,4801215

3,4801215

add a comment | 

add a comment | 

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