Prove logarithmic identity [closed]

1

$begingroup$

I have to prove that $a^{log_b(x)} = x^{log_b(a)}$ without using the base change rule. How might I go about doing this?

logarithms proof-explanation

share|cite|improve this question

edited Jan 15 at 0:56

David G. Stork

11k41432

asked Jan 15 at 0:49

csStudentcsStudent

82

$endgroup$

closed as off-topic by Xander Henderson, Lord Shark the Unknown, Adrian Keister, Eevee Trainer, Cesareo Jan 15 at 2:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, Adrian Keister, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

1

$begingroup$

I have to prove that $a^{log_b(x)} = x^{log_b(a)}$ without using the base change rule. How might I go about doing this?

logarithms proof-explanation

share|cite|improve this question

edited Jan 15 at 0:56

David G. Stork

11k41432

asked Jan 15 at 0:49

csStudentcsStudent

82

$endgroup$

closed as off-topic by Xander Henderson, Lord Shark the Unknown, Adrian Keister, Eevee Trainer, Cesareo Jan 15 at 2:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, Adrian Keister, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

1

1

1

$begingroup$

I have to prove that $a^{log_b(x)} = x^{log_b(a)}$ without using the base change rule. How might I go about doing this?

logarithms proof-explanation

share|cite|improve this question

edited Jan 15 at 0:56

David G. Stork

11k41432

asked Jan 15 at 0:49

csStudentcsStudent

82

$endgroup$

I have to prove that $a^{log_b(x)} = x^{log_b(a)}$ without using the base change rule. How might I go about doing this?

logarithms proof-explanation

logarithms proof-explanation

share|cite|improve this question

edited Jan 15 at 0:56

David G. Stork

11k41432

asked Jan 15 at 0:49

csStudentcsStudent

82

share|cite|improve this question

edited Jan 15 at 0:56

David G. Stork

11k41432

asked Jan 15 at 0:49

csStudentcsStudent

82

share|cite|improve this question

share|cite|improve this question

edited Jan 15 at 0:56

David G. Stork

11k41432

edited Jan 15 at 0:56

David G. Stork

11k41432

edited Jan 15 at 0:56

David G. Stork

11k41432

11k41432

asked Jan 15 at 0:49

csStudentcsStudent

82

asked Jan 15 at 0:49

csStudentcsStudent

82

asked Jan 15 at 0:49

csStudentcsStudent

82

82

closed as off-topic by Xander Henderson, Lord Shark the Unknown, Adrian Keister, Eevee Trainer, Cesareo Jan 15 at 2:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, Adrian Keister, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Xander Henderson, Lord Shark the Unknown, Adrian Keister, Eevee Trainer, Cesareo Jan 15 at 2:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, Adrian Keister, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

1

$begingroup$

Hints:

• take logarithms base $b$ of both sides




• $log_bleft(c^dright) = d, log_b(c)$

share|cite|improve this answer

answered Jan 15 at 0:51

HenryHenry

100k480166

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah that makes sense. I should have looked at log properties closer and thought about it more.
  $endgroup$
  – csStudent
  Jan 15 at 1:01
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $log_bx=y,x=b^y$

$x^{log_ba}=(b^{log_ba})^y=a^y$

The last identity can be established by setting $log_ba=z,a=b^z=b^{log_ba}$

share|cite|improve this answer

answered Jan 15 at 1:18

lab bhattacharjeelab bhattacharjee

226k15157275

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Hints:

• take logarithms base $b$ of both sides




• $log_bleft(c^dright) = d, log_b(c)$

share|cite|improve this answer

answered Jan 15 at 0:51

HenryHenry

100k480166

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah that makes sense. I should have looked at log properties closer and thought about it more.
  $endgroup$
  – csStudent
  Jan 15 at 1:01
  

  
  
  
  

add a comment | 

1

$begingroup$

Hints:

• take logarithms base $b$ of both sides




• $log_bleft(c^dright) = d, log_b(c)$

share|cite|improve this answer

answered Jan 15 at 0:51

HenryHenry

100k480166

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah that makes sense. I should have looked at log properties closer and thought about it more.
  $endgroup$
  – csStudent
  Jan 15 at 1:01
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Hints:

• take logarithms base $b$ of both sides




• $log_bleft(c^dright) = d, log_b(c)$

share|cite|improve this answer

answered Jan 15 at 0:51

HenryHenry

100k480166

$endgroup$

Hints:

• take logarithms base $b$ of both sides




• $log_bleft(c^dright) = d, log_b(c)$

share|cite|improve this answer

answered Jan 15 at 0:51

HenryHenry

100k480166

share|cite|improve this answer

share|cite|improve this answer

answered Jan 15 at 0:51

HenryHenry

100k480166

answered Jan 15 at 0:51

HenryHenry

100k480166

answered Jan 15 at 0:51

HenryHenry

100k480166

100k480166

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah that makes sense. I should have looked at log properties closer and thought about it more.
  $endgroup$
  – csStudent
  Jan 15 at 1:01
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah that makes sense. I should have looked at log properties closer and thought about it more.
  $endgroup$
  – csStudent
  Jan 15 at 1:01
  

  
  
  
  

$begingroup$
Ah that makes sense. I should have looked at log properties closer and thought about it more.
$endgroup$
– csStudent
Jan 15 at 1:01

$begingroup$
Ah that makes sense. I should have looked at log properties closer and thought about it more.
$endgroup$
– csStudent
Jan 15 at 1:01

add a comment | 

0

$begingroup$

Let $log_bx=y,x=b^y$

$x^{log_ba}=(b^{log_ba})^y=a^y$

The last identity can be established by setting $log_ba=z,a=b^z=b^{log_ba}$

share|cite|improve this answer

answered Jan 15 at 1:18

lab bhattacharjeelab bhattacharjee

226k15157275

$endgroup$

add a comment | 

0

$begingroup$

Let $log_bx=y,x=b^y$

$x^{log_ba}=(b^{log_ba})^y=a^y$

The last identity can be established by setting $log_ba=z,a=b^z=b^{log_ba}$

share|cite|improve this answer

answered Jan 15 at 1:18

lab bhattacharjeelab bhattacharjee

226k15157275

$endgroup$

add a comment | 

0

0

0

$begingroup$

Let $log_bx=y,x=b^y$

$x^{log_ba}=(b^{log_ba})^y=a^y$

The last identity can be established by setting $log_ba=z,a=b^z=b^{log_ba}$

share|cite|improve this answer

answered Jan 15 at 1:18

lab bhattacharjeelab bhattacharjee

226k15157275

$endgroup$

Let $log_bx=y,x=b^y$

$x^{log_ba}=(b^{log_ba})^y=a^y$

The last identity can be established by setting $log_ba=z,a=b^z=b^{log_ba}$

share|cite|improve this answer

answered Jan 15 at 1:18

lab bhattacharjeelab bhattacharjee

226k15157275

share|cite|improve this answer

share|cite|improve this answer

answered Jan 15 at 1:18

lab bhattacharjeelab bhattacharjee

226k15157275

answered Jan 15 at 1:18

lab bhattacharjeelab bhattacharjee

226k15157275

answered Jan 15 at 1:18

lab bhattacharjeelab bhattacharjee

226k15157275

226k15157275

add a comment | 

add a comment | 

This page is only for reference, If you need detailed information, please check here