How to parse $2^{1+2^3}$?












3












$begingroup$


I'm fairly comfortable with arithmetic and high school math overall, but I'm having trouble being 100% sure that $2^{large {1+2^3}}$ should be parsed as 2^(1+(2^3)) and not as 2^((1+2)^3), because I've heard that




One must evaluate the base of an exponent before it is raised to the exponent.




As I understand that is actually true and the only way to support the second alternative is to claim that the power is $1+2^3$ (instead of $2^3$ alone) and that its base is $1+2$.



Can anyone confirm that the correct interpretation is 2^(1+(2^3)) and that it yields $512$?






algebra-precalculus arithmetic exponentiation







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $$2^{1+2^3}=2^{1+8}=2^9=512$$
    $endgroup$
    – Dr. Mathva
    Jan 22 at 19:34






  • 2




    $begingroup$
    Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
    $endgroup$
    – Shubham Johri
    Jan 22 at 19:36






  • 2




    $begingroup$
    You are correct. Exponents take precedence over addition.
    $endgroup$
    – Wojowu
    Jan 22 at 20:05






  • 1




    $begingroup$
    This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
    $endgroup$
    – Git Gud
    Jan 23 at 9:09










  • $begingroup$
    Thank you everyone for all your comments.
    $endgroup$
    – Sahil Sunny
    Jan 25 at 6:57
















3












$begingroup$


I'm fairly comfortable with arithmetic and high school math overall, but I'm having trouble being 100% sure that $2^{large {1+2^3}}$ should be parsed as 2^(1+(2^3)) and not as 2^((1+2)^3), because I've heard that




One must evaluate the base of an exponent before it is raised to the exponent.




As I understand that is actually true and the only way to support the second alternative is to claim that the power is $1+2^3$ (instead of $2^3$ alone) and that its base is $1+2$.



Can anyone confirm that the correct interpretation is 2^(1+(2^3)) and that it yields $512$?






algebra-precalculus arithmetic exponentiation







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $$2^{1+2^3}=2^{1+8}=2^9=512$$
    $endgroup$
    – Dr. Mathva
    Jan 22 at 19:34






  • 2




    $begingroup$
    Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
    $endgroup$
    – Shubham Johri
    Jan 22 at 19:36






  • 2




    $begingroup$
    You are correct. Exponents take precedence over addition.
    $endgroup$
    – Wojowu
    Jan 22 at 20:05






  • 1




    $begingroup$
    This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
    $endgroup$
    – Git Gud
    Jan 23 at 9:09










  • $begingroup$
    Thank you everyone for all your comments.
    $endgroup$
    – Sahil Sunny
    Jan 25 at 6:57














3












3








3





$begingroup$


I'm fairly comfortable with arithmetic and high school math overall, but I'm having trouble being 100% sure that $2^{large {1+2^3}}$ should be parsed as 2^(1+(2^3)) and not as 2^((1+2)^3), because I've heard that




One must evaluate the base of an exponent before it is raised to the exponent.




As I understand that is actually true and the only way to support the second alternative is to claim that the power is $1+2^3$ (instead of $2^3$ alone) and that its base is $1+2$.



Can anyone confirm that the correct interpretation is 2^(1+(2^3)) and that it yields $512$?






algebra-precalculus arithmetic exponentiation







share|cite|improve this question









$endgroup$




I'm fairly comfortable with arithmetic and high school math overall, but I'm having trouble being 100% sure that $2^{large {1+2^3}}$ should be parsed as 2^(1+(2^3)) and not as 2^((1+2)^3), because I've heard that




One must evaluate the base of an exponent before it is raised to the exponent.




As I understand that is actually true and the only way to support the second alternative is to claim that the power is $1+2^3$ (instead of $2^3$ alone) and that its base is $1+2$.



Can anyone confirm that the correct interpretation is 2^(1+(2^3)) and that it yields $512$?






algebra-precalculus arithmetic exponentiation




algebra-precalculus arithmetic exponentiation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 22 at 19:32









Sahil SunnySahil Sunny

183




183








  • 2




    $begingroup$
    $$2^{1+2^3}=2^{1+8}=2^9=512$$
    $endgroup$
    – Dr. Mathva
    Jan 22 at 19:34






  • 2




    $begingroup$
    Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
    $endgroup$
    – Shubham Johri
    Jan 22 at 19:36






  • 2




    $begingroup$
    You are correct. Exponents take precedence over addition.
    $endgroup$
    – Wojowu
    Jan 22 at 20:05






  • 1




    $begingroup$
    This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
    $endgroup$
    – Git Gud
    Jan 23 at 9:09










  • $begingroup$
    Thank you everyone for all your comments.
    $endgroup$
    – Sahil Sunny
    Jan 25 at 6:57














  • 2




    $begingroup$
    $$2^{1+2^3}=2^{1+8}=2^9=512$$
    $endgroup$
    – Dr. Mathva
    Jan 22 at 19:34






  • 2




    $begingroup$
    Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
    $endgroup$
    – Shubham Johri
    Jan 22 at 19:36






  • 2




    $begingroup$
    You are correct. Exponents take precedence over addition.
    $endgroup$
    – Wojowu
    Jan 22 at 20:05






  • 1




    $begingroup$
    This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
    $endgroup$
    – Git Gud
    Jan 23 at 9:09










  • $begingroup$
    Thank you everyone for all your comments.
    $endgroup$
    – Sahil Sunny
    Jan 25 at 6:57








2




2




$begingroup$
$$2^{1+2^3}=2^{1+8}=2^9=512$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:34




$begingroup$
$$2^{1+2^3}=2^{1+8}=2^9=512$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:34




2




2




$begingroup$
Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
$endgroup$
– Shubham Johri
Jan 22 at 19:36




$begingroup$
Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
$endgroup$
– Shubham Johri
Jan 22 at 19:36




2




2




$begingroup$
You are correct. Exponents take precedence over addition.
$endgroup$
– Wojowu
Jan 22 at 20:05




$begingroup$
You are correct. Exponents take precedence over addition.
$endgroup$
– Wojowu
Jan 22 at 20:05




1




1




$begingroup$
This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
$endgroup$
– Git Gud
Jan 23 at 9:09




$begingroup$
This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
$endgroup$
– Git Gud
Jan 23 at 9:09












$begingroup$
Thank you everyone for all your comments.
$endgroup$
– Sahil Sunny
Jan 25 at 6:57




$begingroup$
Thank you everyone for all your comments.
$endgroup$
– Sahil Sunny
Jan 25 at 6:57










1 Answer
1






active

oldest

votes


















4












$begingroup$

Yes, your work is indeed correct.



Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.






share|cite|improve this answer











$endgroup$













    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%2f3083603%2fhow-to-parse-2123%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Yes, your work is indeed correct.



    Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



    This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



    Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      Yes, your work is indeed correct.



      Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



      This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



      Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        Yes, your work is indeed correct.



        Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



        This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



        Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.






        share|cite|improve this answer











        $endgroup$



        Yes, your work is indeed correct.



        Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



        This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



        Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 22 at 22:56

























        answered Jan 22 at 19:47









        jordan_glenjordan_glen

        1




        1






























            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%2f3083603%2fhow-to-parse-2123%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

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
            Read more

            Antonio Litta Visconti Arese

            Image
            Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
            Read more