Middle School Log question












1














$$left(frac 43right)^x=frac{8sqrt3}{9}$$



I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.










share|cite|improve this question




















  • 3




    If you have an acceptable answer, accept it by clicking the check mark on that answer.
    – Sean Roberson
    1 hour ago
















1














$$left(frac 43right)^x=frac{8sqrt3}{9}$$



I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.










share|cite|improve this question




















  • 3




    If you have an acceptable answer, accept it by clicking the check mark on that answer.
    – Sean Roberson
    1 hour ago














1












1








1







$$left(frac 43right)^x=frac{8sqrt3}{9}$$



I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.










share|cite|improve this question















$$left(frac 43right)^x=frac{8sqrt3}{9}$$



I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.







logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Rócherz

2,7362721




2,7362721










asked 1 hour ago









Toylatte

113




113








  • 3




    If you have an acceptable answer, accept it by clicking the check mark on that answer.
    – Sean Roberson
    1 hour ago














  • 3




    If you have an acceptable answer, accept it by clicking the check mark on that answer.
    – Sean Roberson
    1 hour ago








3




3




If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
1 hour ago




If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
1 hour ago










2 Answers
2






active

oldest

votes


















6














I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$

Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.






share|cite|improve this answer








New contributor




ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • Thanks it helped
    – Toylatte
    1 hour ago










  • @Toylatte You are welcome!
    – ImNotTheGuy
    1 hour ago



















3














$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



Hence $x=1.5$.



Alternatively, taking logarithm on both sides.






share|cite|improve this answer





















    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%2f3052593%2fmiddle-school-log-question%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









    6














    I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
    $$
    left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
    $$

    Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



    Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.






    share|cite|improve this answer








    New contributor




    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


















    • Thanks it helped
      – Toylatte
      1 hour ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      1 hour ago
















    6














    I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
    $$
    left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
    $$

    Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



    Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.






    share|cite|improve this answer








    New contributor




    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


















    • Thanks it helped
      – Toylatte
      1 hour ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      1 hour ago














    6












    6








    6






    I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
    $$
    left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
    $$

    Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



    Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.






    share|cite|improve this answer








    New contributor




    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
    $$
    left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
    $$

    Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



    Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.







    share|cite|improve this answer








    New contributor




    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    share|cite|improve this answer



    share|cite|improve this answer






    New contributor




    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    answered 1 hour ago









    ImNotTheGuy

    3825




    3825




    New contributor




    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





    New contributor





    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.












    • Thanks it helped
      – Toylatte
      1 hour ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      1 hour ago


















    • Thanks it helped
      – Toylatte
      1 hour ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      1 hour ago
















    Thanks it helped
    – Toylatte
    1 hour ago




    Thanks it helped
    – Toylatte
    1 hour ago












    @Toylatte You are welcome!
    – ImNotTheGuy
    1 hour ago




    @Toylatte You are welcome!
    – ImNotTheGuy
    1 hour ago











    3














    $$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



    Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



    Hence $x=1.5$.



    Alternatively, taking logarithm on both sides.






    share|cite|improve this answer


























      3














      $$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



      Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



      Hence $x=1.5$.



      Alternatively, taking logarithm on both sides.






      share|cite|improve this answer
























        3












        3








        3






        $$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



        Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



        Hence $x=1.5$.



        Alternatively, taking logarithm on both sides.






        share|cite|improve this answer












        $$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



        Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



        Hence $x=1.5$.



        Alternatively, taking logarithm on both sides.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Siong Thye Goh

        98.8k1464116




        98.8k1464116






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3052593%2fmiddle-school-log-question%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

            Quarter-circle Tiles

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Mont Emei