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






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
    33 mins ago










  • $begingroup$
    Did you mean $x=mathbf 2^{-2/3}$ ?
    $endgroup$
    – J. W. Tanner
    18 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






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
    33 mins ago










  • $begingroup$
    Did you mean $x=mathbf 2^{-2/3}$ ?
    $endgroup$
    – J. W. Tanner
    18 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






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 31 mins ago









Eevee Trainer

7,71521338




7,71521338










asked 37 mins ago









KevinKevin

396




396












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










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




















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










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


















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




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












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






$begingroup$
Did you mean $x=mathbf 2^{-2/3}$ ?
$endgroup$
– J. W. Tanner
18 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
      22 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
      8 mins 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 34 mins ago









        Eevee TrainerEevee Trainer

        7,71521338




        7,71521338























            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
              22 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
              8 mins 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
              22 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
              8 mins 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 28 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
              22 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
              8 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
              22 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
              8 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
            22 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
            22 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
            8 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
            8 mins 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

            サソリ

            Image
            この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2019年1月 ） 「 サソリ 」のその他の用法については「サソリ (曖昧さ回避)」をご覧ください。 「 スコーピオン 」と「 スコルピオン 」はこの項目へ転送されています。その他の用法については「スコーピオン」をご覧ください。 サソリ目 生息年代: 430–0 Ma PreЄ Є O S D C P T J K Pg N シルル紀前期-現世 ダイオウサソリ（ Pandinus Imperator ） 分類 界 : 動物界 Animalia 門 : 節足動物門 Arthropoda 亜門 : 鋏角亜門 Chelicerata 綱 : クモガタ綱 Arachnida 目 : サソリ目 Scorpiones 学名 Scorpiones C. L. Koch, 1837 英名 Scorpion 上科 Pseudochactoidea キョクトウサソリ上科 Buthoidea Chaeriloidea Chactoidea Iuroidea Scorpionoidea サソリ （ サソリ類 、 蠍 、 蝎 、 scorpion ）は、鋏角亜門・クモガタ綱の サソリ目 （ Scorpiones ）に属する節足動物の総称である。体の前端に鋏型の触肢、後端に毒針を有する捕食者である。 1700種以上を含め、最古の化石記録は4億3千万年前のシルル紀まで遡る。有毒生物として広く知られるが、人間の命にかかわる毒を持つものはごく一部の種類に過ぎない。 目次 1 形態 1.1 前体 1.2 後体 1.2.1 中体 1.2.2 終体 2 雌雄 3 生態 3.1 天敵 3.2 生殖 4 分布 5 毒 6 蛍光現象 7 分類
            Read more

            広島県道265号伴広島線

            Image
            一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
            Read more

            Setup Asymptote in Texstudio

            Image
            0 Good time of the day! I am trying to compile the following file: documentclass[12pt]{article} usepackage[inline]{asymptote} begin{document} Here is a venn diagram produced with Asymptote, drawn to width 4cm: defA{A} defB{V{B}} begin{figure} begin{center} begin{asy} size(4cm,0); pen colour1=red; pen colour2=green; pair z0=(0,0); pair z1=(-1,0); pair z2=(1,0); real r=1.5; path c1=circle(z1,r); path c2=circle(z2,r); fill(c1,colour1); fill(c2,colour2); picture intersection=new picture; fill(intersection,c1,colour1+colour2); clip(intersection,c2); add(intersection); draw(c1); draw(c2); //draw("$A$",box,z1)
            Read more