Solving a radical function [closed]
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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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closed as off-topic by Eevee Trainer, choco_addicted, Saad, KM101, RRL Jan 5 at 7:45
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If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
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
$endgroup$
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.
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no, it should be 6^(1/6) like raising 6 to the 1/6 power
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– Ryan smith
Jan 5 at 3:31
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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)
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– Ross Millikan
Jan 5 at 3:42
add a comment |
$begingroup$
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
$endgroup$
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
edited Jan 5 at 3:38
Chris Custer
11.5k3824
11.5k3824
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
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Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$
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add a comment |
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$frac{6^1}{6}$ 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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add a comment |
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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 $b=log_xfrac{6^{frac16}}{15}+8$.
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$
$endgroup$
add a comment |
$begingroup$
Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$
$endgroup$
add a comment |
$begingroup$
Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$
$endgroup$
Presumably you are given $x$, so take the base $x$ log of both sides. The right is $b+log_x2$
answered Jan 5 at 3:31
Ross MillikanRoss Millikan
293k23197371
293k23197371
add a comment |
add a comment |
$begingroup$
$frac{6^1}{6}$ 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 $b=log_{x}{frac{2x^8}{30}}$ .
$endgroup$
add a comment |
$begingroup$
$frac{6^1}{6}$ 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 $b=log_{x}{frac{2x^8}{30}}$ .
$endgroup$
add a comment |
$begingroup$
$frac{6^1}{6}$ 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 $b=log_{x}{frac{2x^8}{30}}$ .
$endgroup$
$frac{6^1}{6}$ 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 $b=log_{x}{frac{2x^8}{30}}$ .
answered Jan 5 at 3:35
GnumbertesterGnumbertester
37518
37518
add a comment |
add a comment |
$begingroup$
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 $b=log_xfrac{6^{frac16}}{15}+8$.
$endgroup$
add a comment |
$begingroup$
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 $b=log_xfrac{6^{frac16}}{15}+8$.
$endgroup$
add a comment |
$begingroup$
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 $b=log_xfrac{6^{frac16}}{15}+8$.
$endgroup$
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 $b=log_xfrac{6^{frac16}}{15}+8$.
answered Jan 5 at 4:03
Chris CusterChris Custer
11.5k3824
11.5k3824
add a comment |
add a comment |
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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