Help find my computational error for logarithms












2












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I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?



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algebra-precalculus logarithms







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  • $begingroup$
    $frac 1 {log x}$ is not generally the same as $log frac 1 x $
    $endgroup$
    – J. W. Tanner
    52 mins ago










  • $begingroup$
    Did you mean $x=mathbf 2^{-2/3}$ ?
    $endgroup$
    – J. W. Tanner
    37 mins ago


















2












$begingroup$


I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?



enter image description here






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    $frac 1 {log x}$ is not generally the same as $log frac 1 x $
    $endgroup$
    – J. W. Tanner
    52 mins ago










  • $begingroup$
    Did you mean $x=mathbf 2^{-2/3}$ ?
    $endgroup$
    – J. W. Tanner
    37 mins ago
















2












2








2


1



$begingroup$


I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?



enter image description here






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$




I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?



enter image description here






algebra-precalculus logarithms




algebra-precalculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 50 mins ago









Eevee Trainer

7,72621338




7,72621338










asked 56 mins ago









KevinKevin

446




446












  • $begingroup$
    $frac 1 {log x}$ is not generally the same as $log frac 1 x $
    $endgroup$
    – J. W. Tanner
    52 mins ago










  • $begingroup$
    Did you mean $x=mathbf 2^{-2/3}$ ?
    $endgroup$
    – J. W. Tanner
    37 mins ago




















  • $begingroup$
    $frac 1 {log x}$ is not generally the same as $log frac 1 x $
    $endgroup$
    – J. W. Tanner
    52 mins ago










  • $begingroup$
    Did you mean $x=mathbf 2^{-2/3}$ ?
    $endgroup$
    – J. W. Tanner
    37 mins ago


















$begingroup$
$frac 1 {log x}$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
52 mins ago




$begingroup$
$frac 1 {log x}$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
52 mins ago












$begingroup$
Did you mean $x=mathbf 2^{-2/3}$ ?
$endgroup$
– J. W. Tanner
37 mins ago






$begingroup$
Did you mean $x=mathbf 2^{-2/3}$ ?
$endgroup$
– J. W. Tanner
37 mins ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Note that



$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$



For example, take $x = 4$. Then



$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$



but



$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$



This is where your error lies.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    To get the correct answer, let $L=log_2(x).$



    Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$



    Multiply by $6L$ to get $$3L^2-6=7L.$$



    Thus $$3L^2-7L-6=0$$



    or $$(3L+2)(L-3)=0.$$



    Can you take it from here?






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
      $endgroup$
      – Eevee Trainer
      41 mins ago










    • $begingroup$
      I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
      $endgroup$
      – Kevin
      26 mins ago











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Note that



    $$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$



    For example, take $x = 4$. Then



    $$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$



    but



    $$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$



    This is where your error lies.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Note that



      $$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$



      For example, take $x = 4$. Then



      $$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$



      but



      $$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$



      This is where your error lies.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Note that



        $$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$



        For example, take $x = 4$. Then



        $$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$



        but



        $$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$



        This is where your error lies.






        share|cite|improve this answer









        $endgroup$



        Note that



        $$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$



        For example, take $x = 4$. Then



        $$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$



        but



        $$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$



        This is where your error lies.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 53 mins ago









        Eevee TrainerEevee Trainer

        7,72621338




        7,72621338























            3












            $begingroup$

            To get the correct answer, let $L=log_2(x).$



            Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$



            Multiply by $6L$ to get $$3L^2-6=7L.$$



            Thus $$3L^2-7L-6=0$$



            or $$(3L+2)(L-3)=0.$$



            Can you take it from here?






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
              $endgroup$
              – Eevee Trainer
              41 mins ago










            • $begingroup$
              I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
              $endgroup$
              – Kevin
              26 mins ago
















            3












            $begingroup$

            To get the correct answer, let $L=log_2(x).$



            Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$



            Multiply by $6L$ to get $$3L^2-6=7L.$$



            Thus $$3L^2-7L-6=0$$



            or $$(3L+2)(L-3)=0.$$



            Can you take it from here?






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
              $endgroup$
              – Eevee Trainer
              41 mins ago










            • $begingroup$
              I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
              $endgroup$
              – Kevin
              26 mins ago














            3












            3








            3





            $begingroup$

            To get the correct answer, let $L=log_2(x).$



            Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$



            Multiply by $6L$ to get $$3L^2-6=7L.$$



            Thus $$3L^2-7L-6=0$$



            or $$(3L+2)(L-3)=0.$$



            Can you take it from here?






            share|cite|improve this answer









            $endgroup$



            To get the correct answer, let $L=log_2(x).$



            Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$



            Multiply by $6L$ to get $$3L^2-6=7L.$$



            Thus $$3L^2-7L-6=0$$



            or $$(3L+2)(L-3)=0.$$



            Can you take it from here?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 47 mins ago









            J. W. TannerJ. W. Tanner

            3,0701320




            3,0701320








            • 1




              $begingroup$
              Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
              $endgroup$
              – Eevee Trainer
              41 mins ago










            • $begingroup$
              I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
              $endgroup$
              – Kevin
              26 mins ago














            • 1




              $begingroup$
              Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
              $endgroup$
              – Eevee Trainer
              41 mins ago










            • $begingroup$
              I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
              $endgroup$
              – Kevin
              26 mins ago








            1




            1




            $begingroup$
            Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
            $endgroup$
            – Eevee Trainer
            41 mins ago




            $begingroup$
            Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
            $endgroup$
            – Eevee Trainer
            41 mins ago












            $begingroup$
            I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
            $endgroup$
            – Kevin
            26 mins ago




            $begingroup$
            I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
            $endgroup$
            – Kevin
            26 mins ago


















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