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










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
    32 mins ago
















3












$begingroup$


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



enter image description here










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
    32 mins 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










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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 20 mins ago









user21820

39.4k543155




39.4k543155










asked 2 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
    32 mins 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
    32 mins 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
32 mins 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
32 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









    $endgroup$









    • 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





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3143253%2fhelp-with-frac12-log-2-x-frac1-log-2-x-frac76%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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

        7,67821338




        7,67821338























            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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3143253%2fhelp-with-frac12-log-2-x-frac1-log-2-x-frac76%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            數位音樂下載

            When can things happen in Etherscan, such as the picture below?

            格利澤436b