Solving a radical function [closed]

-1

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I already did the previous algebra and i'm on this one step that I can't seem to get past.

$$6^{frac16}cdotfrac{2x^8}{15}=2x^b$$ solve for b
thank-you

radical-equations

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edited Jan 5 at 3:38

Chris Custer

11.5k3824

asked Jan 5 at 3:25

Ryan smithRyan smith

4

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closed as off-topic by Eevee Trainer, choco_addicted, Saad, KM101, RRL Jan 5 at 7:45

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." – Eevee Trainer, choco_addicted, Saad, KM101, RRL

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

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  $begingroup$
  no, it should be 6^(1/6) like raising 6 to the 1/6 power
  $endgroup$
  – Ryan smith
  Jan 5 at 3:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you should have supplied parentheses to say that. Exponentiation takes precedence over division, so 6^1/6 is $(6^1)/6$ It should be 6^(1/6)
  $endgroup$
  – Ross Millikan
  Jan 5 at 3:42
  
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

I already did the previous algebra and i'm on this one step that I can't seem to get past.

$$6^{frac16}cdotfrac{2x^8}{15}=2x^b$$ solve for b
thank-you

radical-equations

share|cite|improve this question

edited Jan 5 at 3:38

Chris Custer

11.5k3824

asked Jan 5 at 3:25

Ryan smithRyan smith

4

$endgroup$

closed as off-topic by Eevee Trainer, choco_addicted, Saad, KM101, RRL Jan 5 at 7:45

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." – Eevee Trainer, choco_addicted, Saad, KM101, RRL

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  no, it should be 6^(1/6) like raising 6 to the 1/6 power
  $endgroup$
  – Ryan smith
  Jan 5 at 3:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you should have supplied parentheses to say that. Exponentiation takes precedence over division, so 6^1/6 is $(6^1)/6$ It should be 6^(1/6)
  $endgroup$
  – Ross Millikan
  Jan 5 at 3:42
  
  
  

  
  
  
  

add a comment | 

-1

-1

-1

$begingroup$

I already did the previous algebra and i'm on this one step that I can't seem to get past.

$$6^{frac16}cdotfrac{2x^8}{15}=2x^b$$ solve for b
thank-you

radical-equations

share|cite|improve this question

edited Jan 5 at 3:38

Chris Custer

11.5k3824

asked Jan 5 at 3:25

Ryan smithRyan smith

4

$endgroup$

I already did the previous algebra and i'm on this one step that I can't seem to get past.

$$6^{frac16}cdotfrac{2x^8}{15}=2x^b$$ solve for b
thank-you

radical-equations

radical-equations

share|cite|improve this question

edited Jan 5 at 3:38

Chris Custer

11.5k3824

asked Jan 5 at 3:25

Ryan smithRyan smith

4

share|cite|improve this question

edited Jan 5 at 3:38

Chris Custer

11.5k3824

asked Jan 5 at 3:25

Ryan smithRyan smith

4

share|cite|improve this question

share|cite|improve this question

edited Jan 5 at 3:38

Chris Custer

11.5k3824

edited Jan 5 at 3:38

Chris Custer

11.5k3824

edited Jan 5 at 3:38

Chris Custer

11.5k3824

11.5k3824

asked Jan 5 at 3:25

Ryan smithRyan smith

4

asked Jan 5 at 3:25

Ryan smithRyan smith

4

asked Jan 5 at 3:25

Ryan smithRyan smith

4

4

closed as off-topic by Eevee Trainer, choco_addicted, Saad, KM101, RRL Jan 5 at 7:45

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." – Eevee Trainer, choco_addicted, Saad, KM101, RRL

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

closed as off-topic by Eevee Trainer, choco_addicted, Saad, KM101, RRL Jan 5 at 7:45

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." – Eevee Trainer, choco_addicted, Saad, KM101, RRL

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  no, it should be 6^(1/6) like raising 6 to the 1/6 power
  $endgroup$
  – Ryan smith
  Jan 5 at 3:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you should have supplied parentheses to say that. Exponentiation takes precedence over division, so 6^1/6 is $(6^1)/6$ It should be 6^(1/6)
  $endgroup$
  – Ross Millikan
  Jan 5 at 3:42
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  no, it should be 6^(1/6) like raising 6 to the 1/6 power
  $endgroup$
  – Ryan smith
  Jan 5 at 3:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you should have supplied parentheses to say that. Exponentiation takes precedence over division, so 6^1/6 is $(6^1)/6$ It should be 6^(1/6)
  $endgroup$
  – Ross Millikan
  Jan 5 at 3:42
  
  
  

  
  
  
  

$begingroup$
no, it should be 6^(1/6) like raising 6 to the 1/6 power
$endgroup$
– Ryan smith
Jan 5 at 3:31

$begingroup$
no, it should be 6^(1/6) like raising 6 to the 1/6 power
$endgroup$
– Ryan smith
Jan 5 at 3:31

$begingroup$
Then you should have supplied parentheses to say that. Exponentiation takes precedence over division, so 6^1/6 is $(6^1)/6$ It should be 6^(1/6)
$endgroup$
– Ross Millikan
Jan 5 at 3:42

$begingroup$
Then you should have supplied parentheses to say that. Exponentiation takes precedence over division, so 6^1/6 is $(6^1)/6$ It should be 6^(1/6)
$endgroup$
– Ross Millikan
Jan 5 at 3:42

add a comment | 

3 Answers
3

active

oldest

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1

$begingroup$

Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$

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answered Jan 5 at 3:31

Ross MillikanRoss Millikan

293k23197371

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add a comment | 

0

$begingroup$

$frac{6^1}{6}$ is 1 so we are left with $frac{2x^8}{15}=2x^b$ which is equivalent to $frac{2x^8}{30}=x^b$. Now just rewrite with logaritms, so $b=log_{x}{frac{2x^8}{30}}$ .

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answered Jan 5 at 3:35

GnumbertesterGnumbertester

37518

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add a comment | 

0

$begingroup$

You could isolate $x^b$ first: $$6^{frac 16}frac{x^8}{15}=x^b$$.

Then, to get $b$ out of the exponent, take $log_x$ of both sides.

Get $b= log_x(6^{frac 16}frac{x^8}{15})$ or $b=log_xfrac{6^{frac16}}{15}+8$.

share|cite|improve this answer

answered Jan 5 at 4:03

Chris CusterChris Custer

11.5k3824

$endgroup$

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$

share|cite|improve this answer

answered Jan 5 at 3:31

Ross MillikanRoss Millikan

293k23197371

$endgroup$

add a comment | 

1

$begingroup$

Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$

share|cite|improve this answer

answered Jan 5 at 3:31

Ross MillikanRoss Millikan

293k23197371

$endgroup$

add a comment | 

1

1

1

$begingroup$

Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$

share|cite|improve this answer

answered Jan 5 at 3:31

Ross MillikanRoss Millikan

293k23197371

$endgroup$

Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$

share|cite|improve this answer

answered Jan 5 at 3:31

Ross MillikanRoss Millikan

293k23197371

share|cite|improve this answer

share|cite|improve this answer

answered Jan 5 at 3:31

Ross MillikanRoss Millikan

293k23197371

answered Jan 5 at 3:31

Ross MillikanRoss Millikan

293k23197371

answered Jan 5 at 3:31

Ross MillikanRoss Millikan

293k23197371

293k23197371

add a comment | 

add a comment | 

0

$begingroup$

$frac{6^1}{6}$ is 1 so we are left with $frac{2x^8}{15}=2x^b$ which is equivalent to $frac{2x^8}{30}=x^b$. Now just rewrite with logaritms, so $b=log_{x}{frac{2x^8}{30}}$ .

share|cite|improve this answer

answered Jan 5 at 3:35

GnumbertesterGnumbertester

37518

$endgroup$

add a comment | 

0

$begingroup$

$frac{6^1}{6}$ is 1 so we are left with $frac{2x^8}{15}=2x^b$ which is equivalent to $frac{2x^8}{30}=x^b$. Now just rewrite with logaritms, so $b=log_{x}{frac{2x^8}{30}}$ .

share|cite|improve this answer

answered Jan 5 at 3:35

GnumbertesterGnumbertester

37518

$endgroup$

add a comment | 

0

0

0

$begingroup$

$frac{6^1}{6}$ is 1 so we are left with $frac{2x^8}{15}=2x^b$ which is equivalent to $frac{2x^8}{30}=x^b$. Now just rewrite with logaritms, so $b=log_{x}{frac{2x^8}{30}}$ .

share|cite|improve this answer

answered Jan 5 at 3:35

GnumbertesterGnumbertester

37518

$endgroup$

$frac{6^1}{6}$ is 1 so we are left with $frac{2x^8}{15}=2x^b$ which is equivalent to $frac{2x^8}{30}=x^b$. Now just rewrite with logaritms, so $b=log_{x}{frac{2x^8}{30}}$ .

share|cite|improve this answer

answered Jan 5 at 3:35

GnumbertesterGnumbertester

37518

share|cite|improve this answer

share|cite|improve this answer

answered Jan 5 at 3:35

GnumbertesterGnumbertester

37518

answered Jan 5 at 3:35

GnumbertesterGnumbertester

37518

answered Jan 5 at 3:35

GnumbertesterGnumbertester

37518

37518

add a comment | 

add a comment | 

0

$begingroup$

You could isolate $x^b$ first: $$6^{frac 16}frac{x^8}{15}=x^b$$.

Then, to get $b$ out of the exponent, take $log_x$ of both sides.

Get $b= log_x(6^{frac 16}frac{x^8}{15})$ or $b=log_xfrac{6^{frac16}}{15}+8$.

share|cite|improve this answer

answered Jan 5 at 4:03

Chris CusterChris Custer

11.5k3824

$endgroup$

add a comment | 

0

$begingroup$

You could isolate $x^b$ first: $$6^{frac 16}frac{x^8}{15}=x^b$$.

Then, to get $b$ out of the exponent, take $log_x$ of both sides.

Get $b= log_x(6^{frac 16}frac{x^8}{15})$ or $b=log_xfrac{6^{frac16}}{15}+8$.

share|cite|improve this answer

answered Jan 5 at 4:03

Chris CusterChris Custer

11.5k3824

$endgroup$

add a comment | 

0

0

0

$begingroup$

You could isolate $x^b$ first: $$6^{frac 16}frac{x^8}{15}=x^b$$.

Then, to get $b$ out of the exponent, take $log_x$ of both sides.

Get $b= log_x(6^{frac 16}frac{x^8}{15})$ or $b=log_xfrac{6^{frac16}}{15}+8$.

share|cite|improve this answer

answered Jan 5 at 4:03

Chris CusterChris Custer

11.5k3824

$endgroup$

You could isolate $x^b$ first: $$6^{frac 16}frac{x^8}{15}=x^b$$.

Then, to get $b$ out of the exponent, take $log_x$ of both sides.

Get $b= log_x(6^{frac 16}frac{x^8}{15})$ or $b=log_xfrac{6^{frac16}}{15}+8$.

share|cite|improve this answer

answered Jan 5 at 4:03

Chris CusterChris Custer

11.5k3824

share|cite|improve this answer

share|cite|improve this answer

answered Jan 5 at 4:03

Chris CusterChris Custer

11.5k3824

answered Jan 5 at 4:03

Chris CusterChris Custer

11.5k3824

answered Jan 5 at 4:03

Chris CusterChris Custer

11.5k3824

11.5k3824

add a comment | 

add a comment | 

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