Help find my computational error for logarithms












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











$endgroup$












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










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


















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











$endgroup$












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










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
















1












1








1


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 24 mins ago









Eevee Trainer

7,70521338




7,70521338










asked 30 mins ago









KevinKevin

396




396












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










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




















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










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


















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




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












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






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












2 Answers
2






active

oldest

votes


















2












$begingroup$

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









$endgroup$





















    2












    $begingroup$

    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
      15 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
      1 min 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-find-my-computational-error-for-logarithms%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









    2












    $begingroup$

    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









    $endgroup$


















      2












      $begingroup$

      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









      $endgroup$
















        2












        2








        2





        $begingroup$

        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 27 mins ago









        Eevee TrainerEevee Trainer

        7,70521338




        7,70521338























            2












            $begingroup$

            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
              15 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
              1 min ago
















            2












            $begingroup$

            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
              15 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
              1 min ago














            2












            2








            2





            $begingroup$

            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 21 mins ago









            J. W. TannerJ. W. Tanner

            3,0501320




            3,0501320








            • 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
              15 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
              1 min 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
              15 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
              1 min 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
            15 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
            15 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
            1 min 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
            1 min 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-find-my-computational-error-for-logarithms%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

            Knooppunt Holsloot

            Altaar (religie)

            Gregoriusmis