Help with $frac12 log_2 x - frac1{log_2 x} = frac76$












3












$begingroup$


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
    2 hours ago










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












  • $begingroup$
    Welcome to Math SE, and thanks for providing your own thoughts on the problem. But next time use MathJax as required. It's not only much neater than handwriting, it's searchable.
    $endgroup$
    – user21820
    1 hour ago
















3












$begingroup$


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
    2 hours ago










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












  • $begingroup$
    Welcome to Math SE, and thanks for providing your own thoughts on the problem. But next time use MathJax as required. It's not only much neater than handwriting, it's searchable.
    $endgroup$
    – user21820
    1 hour ago














3












3








3


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 51 mins ago









user21820

39.4k543155




39.4k543155










asked 3 hours ago









KevinKevin

496




496












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










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












  • $begingroup$
    Welcome to Math SE, and thanks for providing your own thoughts on the problem. But next time use MathJax as required. It's not only much neater than handwriting, it's searchable.
    $endgroup$
    – user21820
    1 hour ago


















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










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












  • $begingroup$
    Welcome to Math SE, and thanks for providing your own thoughts on the problem. But next time use MathJax as required. It's not only much neater than handwriting, it's searchable.
    $endgroup$
    – user21820
    1 hour ago
















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




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












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






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














$begingroup$
Welcome to Math SE, and thanks for providing your own thoughts on the problem. But next time use MathJax as required. It's not only much neater than handwriting, it's searchable.
$endgroup$
– user21820
1 hour ago




$begingroup$
Welcome to Math SE, and thanks for providing your own thoughts on the problem. But next time use MathJax as required. It's not only much neater than handwriting, it's searchable.
$endgroup$
– user21820
1 hour 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









    $endgroup$









    • 2




      $begingroup$
      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
      2 hours 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
      2 hours 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 2 hours ago









        Eevee TrainerEevee Trainer

        7,67721338




        7,67721338























            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









            $endgroup$









            • 2




              $begingroup$
              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
              2 hours 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
              2 hours 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









            $endgroup$









            • 2




              $begingroup$
              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
              2 hours 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
              2 hours 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 2 hours ago









            J. W. TannerJ. W. Tanner

            3,0681320




            3,0681320








            • 2




              $begingroup$
              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
              2 hours 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
              2 hours ago














            • 2




              $begingroup$
              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
              2 hours 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
              2 hours ago








            2




            2




            $begingroup$
            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
            2 hours 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
            2 hours 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
            2 hours 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
            2 hours ago


















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