How to find x in $x^{log _{2}x}>16$

0

$begingroup$

$$x^{log _{2} x}>16$$
What I have done is :

1. Take log fo both sides


2. Then I don't know to do what!
   Please help me if it is possible.
   Hint me about path through solving it.

logarithms

share|cite|improve this question

edited Jan 27 at 17:57

Thomas Shelby

4,4292726

asked Jan 27 at 17:55

Mr invisibleMr invisible

193

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just take $log_2$ on both sides.
  $endgroup$
  – Tito Eliatron
  Jan 27 at 17:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
  $endgroup$
  – robjohn♦
  Jan 27 at 18:15
  

  
  
  
  

add a comment | 

0

$begingroup$

$$x^{log _{2} x}>16$$
What I have done is :

1. Take log fo both sides


2. Then I don't know to do what!
   Please help me if it is possible.
   Hint me about path through solving it.

logarithms

share|cite|improve this question

edited Jan 27 at 17:57

Thomas Shelby

4,4292726

asked Jan 27 at 17:55

Mr invisibleMr invisible

193

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just take $log_2$ on both sides.
  $endgroup$
  – Tito Eliatron
  Jan 27 at 17:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
  $endgroup$
  – robjohn♦
  Jan 27 at 18:15
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

$$x^{log _{2} x}>16$$
What I have done is :

1. Take log fo both sides


2. Then I don't know to do what!
   Please help me if it is possible.
   Hint me about path through solving it.

logarithms

share|cite|improve this question

edited Jan 27 at 17:57

Thomas Shelby

4,4292726

asked Jan 27 at 17:55

Mr invisibleMr invisible

193

$endgroup$

$$x^{log _{2} x}>16$$
What I have done is :

1. Take log fo both sides


2. Then I don't know to do what!
   Please help me if it is possible.
   Hint me about path through solving it.

logarithms

logarithms

share|cite|improve this question

edited Jan 27 at 17:57

Thomas Shelby

4,4292726

asked Jan 27 at 17:55

Mr invisibleMr invisible

193

share|cite|improve this question

edited Jan 27 at 17:57

Thomas Shelby

4,4292726

asked Jan 27 at 17:55

Mr invisibleMr invisible

193

share|cite|improve this question

share|cite|improve this question

edited Jan 27 at 17:57

Thomas Shelby

4,4292726

edited Jan 27 at 17:57

Thomas Shelby

4,4292726

edited Jan 27 at 17:57

Thomas Shelby

4,4292726

4,4292726

asked Jan 27 at 17:55

Mr invisibleMr invisible

193

asked Jan 27 at 17:55

Mr invisibleMr invisible

193

asked Jan 27 at 17:55

Mr invisibleMr invisible

193

193

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just take $log_2$ on both sides.
  $endgroup$
  – Tito Eliatron
  Jan 27 at 17:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
  $endgroup$
  – robjohn♦
  Jan 27 at 18:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just take $log_2$ on both sides.
  $endgroup$
  – Tito Eliatron
  Jan 27 at 17:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
  $endgroup$
  – robjohn♦
  Jan 27 at 18:15
  

  
  
  
  

1

1

$begingroup$
Just take $log_2$ on both sides.
$endgroup$
– Tito Eliatron
Jan 27 at 17:59

$begingroup$
Just take $log_2$ on both sides.
$endgroup$
– Tito Eliatron
Jan 27 at 17:59

$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Jan 27 at 18:15

$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Jan 27 at 18:15

add a comment | 

3 Answers
3

active

oldest

votes

1

$begingroup$

Taking $log_2$ on both sides, we get:

$$log_2(x^{log_2(x)})>log_2(16)$$
By the power rule: $$log(a^b)=blog(a)$$ we get:

$$(log_2(x))^2>log_2(16)$$

Evaluate $log_2(16)$ and go from there.

share|cite|improve this answer

answered Jan 27 at 18:05

Rhys HughesRhys Hughes

7,0801630

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Would you explain that how you used that power rule in second step?
  $endgroup$
  – Mr invisible
  Jan 27 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I brought the power on the LHS down, so: $$log_2(x^{log_2(x)})=log_2(x)log_2(x)=(log_2(x))^2$$
  $endgroup$
  – Rhys Hughes
  Jan 27 at 18:38
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: Observe
begin{align}
x^{log_2 x} = 2^{(log_2 x)^2}.
end{align}

share|cite|improve this answer

answered Jan 27 at 17:59

Jacky ChongJacky Chong

19.4k21129

$endgroup$

add a comment | 

0

$begingroup$

If $log_2x=y, x=2^y$

We need $(2^y)^y>16iff2^{y^2}>2^4implies y^2>4$

$implies(i)$ either $y>2iff x>2^2$

$(ii)$ or $y<-2iff x<2^{-2}$

share|cite|improve this answer

answered Jan 27 at 18:23

lab bhattacharjeelab bhattacharjee

227k15158278

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Taking $log_2$ on both sides, we get:

$$log_2(x^{log_2(x)})>log_2(16)$$
By the power rule: $$log(a^b)=blog(a)$$ we get:

$$(log_2(x))^2>log_2(16)$$

Evaluate $log_2(16)$ and go from there.

share|cite|improve this answer

answered Jan 27 at 18:05

Rhys HughesRhys Hughes

7,0801630

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Would you explain that how you used that power rule in second step?
  $endgroup$
  – Mr invisible
  Jan 27 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I brought the power on the LHS down, so: $$log_2(x^{log_2(x)})=log_2(x)log_2(x)=(log_2(x))^2$$
  $endgroup$
  – Rhys Hughes
  Jan 27 at 18:38
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Taking $log_2$ on both sides, we get:

$$log_2(x^{log_2(x)})>log_2(16)$$
By the power rule: $$log(a^b)=blog(a)$$ we get:

$$(log_2(x))^2>log_2(16)$$

Evaluate $log_2(16)$ and go from there.

share|cite|improve this answer

answered Jan 27 at 18:05

Rhys HughesRhys Hughes

7,0801630

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Would you explain that how you used that power rule in second step?
  $endgroup$
  – Mr invisible
  Jan 27 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I brought the power on the LHS down, so: $$log_2(x^{log_2(x)})=log_2(x)log_2(x)=(log_2(x))^2$$
  $endgroup$
  – Rhys Hughes
  Jan 27 at 18:38
  
  
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Taking $log_2$ on both sides, we get:

$$log_2(x^{log_2(x)})>log_2(16)$$
By the power rule: $$log(a^b)=blog(a)$$ we get:

$$(log_2(x))^2>log_2(16)$$

Evaluate $log_2(16)$ and go from there.

share|cite|improve this answer

answered Jan 27 at 18:05

Rhys HughesRhys Hughes

7,0801630

$endgroup$

Taking $log_2$ on both sides, we get:

$$log_2(x^{log_2(x)})>log_2(16)$$
By the power rule: $$log(a^b)=blog(a)$$ we get:

$$(log_2(x))^2>log_2(16)$$

Evaluate $log_2(16)$ and go from there.

share|cite|improve this answer

answered Jan 27 at 18:05

Rhys HughesRhys Hughes

7,0801630

share|cite|improve this answer

share|cite|improve this answer

answered Jan 27 at 18:05

Rhys HughesRhys Hughes

7,0801630

answered Jan 27 at 18:05

Rhys HughesRhys Hughes

7,0801630

answered Jan 27 at 18:05

Rhys HughesRhys Hughes

7,0801630

7,0801630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Would you explain that how you used that power rule in second step?
  $endgroup$
  – Mr invisible
  Jan 27 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I brought the power on the LHS down, so: $$log_2(x^{log_2(x)})=log_2(x)log_2(x)=(log_2(x))^2$$
  $endgroup$
  – Rhys Hughes
  Jan 27 at 18:38
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Would you explain that how you used that power rule in second step?
  $endgroup$
  – Mr invisible
  Jan 27 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I brought the power on the LHS down, so: $$log_2(x^{log_2(x)})=log_2(x)log_2(x)=(log_2(x))^2$$
  $endgroup$
  – Rhys Hughes
  Jan 27 at 18:38
  
  
  

  
  
  
  

$begingroup$
Would you explain that how you used that power rule in second step?
$endgroup$
– Mr invisible
Jan 27 at 18:15

$begingroup$
Would you explain that how you used that power rule in second step?
$endgroup$
– Mr invisible
Jan 27 at 18:15

$begingroup$
I brought the power on the LHS down, so: $$log_2(x^{log_2(x)})=log_2(x)log_2(x)=(log_2(x))^2$$
$endgroup$
– Rhys Hughes
Jan 27 at 18:38

$begingroup$
I brought the power on the LHS down, so: $$log_2(x^{log_2(x)})=log_2(x)log_2(x)=(log_2(x))^2$$
$endgroup$
– Rhys Hughes
Jan 27 at 18:38

add a comment | 

1

$begingroup$

Hint: Observe
begin{align}
x^{log_2 x} = 2^{(log_2 x)^2}.
end{align}

share|cite|improve this answer

answered Jan 27 at 17:59

Jacky ChongJacky Chong

19.4k21129

$endgroup$

add a comment | 

1

$begingroup$

Hint: Observe
begin{align}
x^{log_2 x} = 2^{(log_2 x)^2}.
end{align}

share|cite|improve this answer

answered Jan 27 at 17:59

Jacky ChongJacky Chong

19.4k21129

$endgroup$

add a comment | 

1

1

1

$begingroup$

Hint: Observe
begin{align}
x^{log_2 x} = 2^{(log_2 x)^2}.
end{align}

share|cite|improve this answer

answered Jan 27 at 17:59

Jacky ChongJacky Chong

19.4k21129

$endgroup$

Hint: Observe
begin{align}
x^{log_2 x} = 2^{(log_2 x)^2}.
end{align}

share|cite|improve this answer

answered Jan 27 at 17:59

Jacky ChongJacky Chong

19.4k21129

share|cite|improve this answer

share|cite|improve this answer

answered Jan 27 at 17:59

Jacky ChongJacky Chong

19.4k21129

answered Jan 27 at 17:59

Jacky ChongJacky Chong

19.4k21129

answered Jan 27 at 17:59

Jacky ChongJacky Chong

19.4k21129

19.4k21129

add a comment | 

add a comment | 

0

$begingroup$

If $log_2x=y, x=2^y$

We need $(2^y)^y>16iff2^{y^2}>2^4implies y^2>4$

$implies(i)$ either $y>2iff x>2^2$

$(ii)$ or $y<-2iff x<2^{-2}$

share|cite|improve this answer

answered Jan 27 at 18:23

lab bhattacharjeelab bhattacharjee

227k15158278

$endgroup$

add a comment | 

0

$begingroup$

If $log_2x=y, x=2^y$

We need $(2^y)^y>16iff2^{y^2}>2^4implies y^2>4$

$implies(i)$ either $y>2iff x>2^2$

$(ii)$ or $y<-2iff x<2^{-2}$

share|cite|improve this answer

answered Jan 27 at 18:23

lab bhattacharjeelab bhattacharjee

227k15158278

$endgroup$

add a comment | 

0

0

0

$begingroup$

If $log_2x=y, x=2^y$

We need $(2^y)^y>16iff2^{y^2}>2^4implies y^2>4$

$implies(i)$ either $y>2iff x>2^2$

$(ii)$ or $y<-2iff x<2^{-2}$

share|cite|improve this answer

answered Jan 27 at 18:23

lab bhattacharjeelab bhattacharjee

227k15158278

$endgroup$

If $log_2x=y, x=2^y$

We need $(2^y)^y>16iff2^{y^2}>2^4implies y^2>4$

$implies(i)$ either $y>2iff x>2^2$

$(ii)$ or $y<-2iff x<2^{-2}$

share|cite|improve this answer

answered Jan 27 at 18:23

lab bhattacharjeelab bhattacharjee

227k15158278

share|cite|improve this answer

share|cite|improve this answer

answered Jan 27 at 18:23

lab bhattacharjeelab bhattacharjee

227k15158278

answered Jan 27 at 18:23

lab bhattacharjeelab bhattacharjee

227k15158278

answered Jan 27 at 18:23

lab bhattacharjeelab bhattacharjee

227k15158278

227k15158278

add a comment | 

add a comment | 

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But avoid …

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Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

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