Solving a radical function [closed]












-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







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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.













  • $begingroup$
    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


















-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











$endgroup$



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
















-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











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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




















  • $begingroup$
    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












3 Answers
3






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









$endgroup$





















    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









    $endgroup$





















      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









      $endgroup$




















        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









        $endgroup$


















          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









          $endgroup$
















            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









            $endgroup$



            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












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 5 at 3:31









            Ross MillikanRoss Millikan

            293k23197371




            293k23197371























                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









                $endgroup$


















                  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









                  $endgroup$
















                    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









                    $endgroup$



                    $frac{6^1}{6}$ 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












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 5 at 3:35









                    GnumbertesterGnumbertester

                    37518




                    37518























                        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









                        $endgroup$


















                          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









                          $endgroup$
















                            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









                            $endgroup$



                            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












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 4:03









                            Chris CusterChris Custer

                            11.5k3824




                            11.5k3824















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