Basic Mathematics, Rules for Multiplication trouble












0














Doing some self study from the text Basic Mathematics by Serge Lang
I ran into an exercise question which I can't seem to wrap my head around.
The question is:



Express the following expressions in the form $2^m3^na^rb^s$ ,where $m,n,r,s$ are positive integers.



$8a^2b^3(27a^4)(2^5ab)$



After some research I found that the final answer is expressed as



$2^83^3a^7b^4$



I've attempted to use distribution as a means of solving it but end up stuck and confused.
I'm entirely lost as to how that answer is derived.






algebra-precalculus exponentiation







share|cite|improve this question





























    0














    Doing some self study from the text Basic Mathematics by Serge Lang
    I ran into an exercise question which I can't seem to wrap my head around.
    The question is:



    Express the following expressions in the form $2^m3^na^rb^s$ ,where $m,n,r,s$ are positive integers.



    $8a^2b^3(27a^4)(2^5ab)$



    After some research I found that the final answer is expressed as



    $2^83^3a^7b^4$



    I've attempted to use distribution as a means of solving it but end up stuck and confused.
    I'm entirely lost as to how that answer is derived.






    algebra-precalculus exponentiation







    share|cite|improve this question



























      0












      0








      0







      Doing some self study from the text Basic Mathematics by Serge Lang
      I ran into an exercise question which I can't seem to wrap my head around.
      The question is:



      Express the following expressions in the form $2^m3^na^rb^s$ ,where $m,n,r,s$ are positive integers.



      $8a^2b^3(27a^4)(2^5ab)$



      After some research I found that the final answer is expressed as



      $2^83^3a^7b^4$



      I've attempted to use distribution as a means of solving it but end up stuck and confused.
      I'm entirely lost as to how that answer is derived.






      algebra-precalculus exponentiation







      share|cite|improve this question















      Doing some self study from the text Basic Mathematics by Serge Lang
      I ran into an exercise question which I can't seem to wrap my head around.
      The question is:



      Express the following expressions in the form $2^m3^na^rb^s$ ,where $m,n,r,s$ are positive integers.



      $8a^2b^3(27a^4)(2^5ab)$



      After some research I found that the final answer is expressed as



      $2^83^3a^7b^4$



      I've attempted to use distribution as a means of solving it but end up stuck and confused.
      I'm entirely lost as to how that answer is derived.






      algebra-precalculus exponentiation




      algebra-precalculus exponentiation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 at 2:49









      timtfj

      977317




      977317










      asked Nov 28 at 23:50









      Reiburn

      1




      1






















          3 Answers
          3






          active

          oldest

          votes


















          4














          The key idea here is that multiplication is both commutative and associative, so we may multiply in any order and in any groupings we wish, as well as "break up" products into any groupings and any order we wish.



          The expression $8a^2b^3(27a^4)(2^5ab)$, by associativity and commutativity, is just the same expression without the parentheses and with the powers of 2 first, then the powers of 3, then the powers of $a$, and finally the powers of $b$. So $8a^2b^3(27a^4)(2^5ab)=8cdot2^5cdot27cdot a^2a^4ab^3b$.



          Using our rules of exponents, we can then condense these products of similar numbers into one big power. So $8cdot2^5cdot27cdot a^2a^4ab^3b = 2^83^3a^7b^4$, which is the answer your research uncovered.






          share|cite|improve this answer





















          • +1 Nicely written, at just the right level of abstraction. I hope the OP appreciates it.
            – Ethan Bolker
            Nov 29 at 0:00










          • Appreciate the response a lot, after testing this method on a few other exercises I believe I understand it clearly.
            – Reiburn
            Nov 29 at 0:01










          • @Reiburn Glad to be of service!
            – JDMan4444
            Nov 29 at 0:03



















          0














          You just apply basic power rules:

          $a^na^m = a^{n+m}$ etc ...


          Let's apply them now to your example:
          $$8a^2b^3(27)a^2b^3a^4ab$$
          $$8 (27) 2^5 a^2 b^3 a^4 ab$$
          $$2^3 2^5 (27) a^7 b^4$$
          $$2^8 3^3 a^7 b^4$$






          share|cite|improve this answer





























            0














            I also have another way to do it. If you distribute the $8 a^2 b^3$ with the $27 a^4$, you will get $216 a^6 b^3$. Then, distribute that value with $2^5 a b$ and you will get $6,912 a^7 b^4$. The number $6,912$ can be written exponentially as $(2^8)(3^3)$. Bring that value down with the $a^7 b^4$, and you get your answer.






            share|cite|improve this answer



















            • 4




              "Distribute"? In my experience, that word usually means something quite different from what you're doing here.
              – David K
              Nov 29 at 0:08












            • By distribute, I really just mean multiply. When you do normal multiplication, you can really think of it is distributing between monomials, which is why it sounds different.
              – Xavier Stanton
              Nov 29 at 0:17






            • 1




              I got what you meant; the benefit of "thinking of" definitions in this way just isn't clear to me.
              – David K
              Nov 29 at 0:21











            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%2f3017928%2fbasic-mathematics-rules-for-multiplication-trouble%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            The key idea here is that multiplication is both commutative and associative, so we may multiply in any order and in any groupings we wish, as well as "break up" products into any groupings and any order we wish.



            The expression $8a^2b^3(27a^4)(2^5ab)$, by associativity and commutativity, is just the same expression without the parentheses and with the powers of 2 first, then the powers of 3, then the powers of $a$, and finally the powers of $b$. So $8a^2b^3(27a^4)(2^5ab)=8cdot2^5cdot27cdot a^2a^4ab^3b$.



            Using our rules of exponents, we can then condense these products of similar numbers into one big power. So $8cdot2^5cdot27cdot a^2a^4ab^3b = 2^83^3a^7b^4$, which is the answer your research uncovered.






            share|cite|improve this answer





















            • +1 Nicely written, at just the right level of abstraction. I hope the OP appreciates it.
              – Ethan Bolker
              Nov 29 at 0:00










            • Appreciate the response a lot, after testing this method on a few other exercises I believe I understand it clearly.
              – Reiburn
              Nov 29 at 0:01










            • @Reiburn Glad to be of service!
              – JDMan4444
              Nov 29 at 0:03
















            4














            The key idea here is that multiplication is both commutative and associative, so we may multiply in any order and in any groupings we wish, as well as "break up" products into any groupings and any order we wish.



            The expression $8a^2b^3(27a^4)(2^5ab)$, by associativity and commutativity, is just the same expression without the parentheses and with the powers of 2 first, then the powers of 3, then the powers of $a$, and finally the powers of $b$. So $8a^2b^3(27a^4)(2^5ab)=8cdot2^5cdot27cdot a^2a^4ab^3b$.



            Using our rules of exponents, we can then condense these products of similar numbers into one big power. So $8cdot2^5cdot27cdot a^2a^4ab^3b = 2^83^3a^7b^4$, which is the answer your research uncovered.






            share|cite|improve this answer





















            • +1 Nicely written, at just the right level of abstraction. I hope the OP appreciates it.
              – Ethan Bolker
              Nov 29 at 0:00










            • Appreciate the response a lot, after testing this method on a few other exercises I believe I understand it clearly.
              – Reiburn
              Nov 29 at 0:01










            • @Reiburn Glad to be of service!
              – JDMan4444
              Nov 29 at 0:03














            4












            4








            4






            The key idea here is that multiplication is both commutative and associative, so we may multiply in any order and in any groupings we wish, as well as "break up" products into any groupings and any order we wish.



            The expression $8a^2b^3(27a^4)(2^5ab)$, by associativity and commutativity, is just the same expression without the parentheses and with the powers of 2 first, then the powers of 3, then the powers of $a$, and finally the powers of $b$. So $8a^2b^3(27a^4)(2^5ab)=8cdot2^5cdot27cdot a^2a^4ab^3b$.



            Using our rules of exponents, we can then condense these products of similar numbers into one big power. So $8cdot2^5cdot27cdot a^2a^4ab^3b = 2^83^3a^7b^4$, which is the answer your research uncovered.






            share|cite|improve this answer












            The key idea here is that multiplication is both commutative and associative, so we may multiply in any order and in any groupings we wish, as well as "break up" products into any groupings and any order we wish.



            The expression $8a^2b^3(27a^4)(2^5ab)$, by associativity and commutativity, is just the same expression without the parentheses and with the powers of 2 first, then the powers of 3, then the powers of $a$, and finally the powers of $b$. So $8a^2b^3(27a^4)(2^5ab)=8cdot2^5cdot27cdot a^2a^4ab^3b$.



            Using our rules of exponents, we can then condense these products of similar numbers into one big power. So $8cdot2^5cdot27cdot a^2a^4ab^3b = 2^83^3a^7b^4$, which is the answer your research uncovered.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 28 at 23:55









            JDMan4444

            23514




            23514












            • +1 Nicely written, at just the right level of abstraction. I hope the OP appreciates it.
              – Ethan Bolker
              Nov 29 at 0:00










            • Appreciate the response a lot, after testing this method on a few other exercises I believe I understand it clearly.
              – Reiburn
              Nov 29 at 0:01










            • @Reiburn Glad to be of service!
              – JDMan4444
              Nov 29 at 0:03


















            • +1 Nicely written, at just the right level of abstraction. I hope the OP appreciates it.
              – Ethan Bolker
              Nov 29 at 0:00










            • Appreciate the response a lot, after testing this method on a few other exercises I believe I understand it clearly.
              – Reiburn
              Nov 29 at 0:01










            • @Reiburn Glad to be of service!
              – JDMan4444
              Nov 29 at 0:03
















            +1 Nicely written, at just the right level of abstraction. I hope the OP appreciates it.
            – Ethan Bolker
            Nov 29 at 0:00




            +1 Nicely written, at just the right level of abstraction. I hope the OP appreciates it.
            – Ethan Bolker
            Nov 29 at 0:00












            Appreciate the response a lot, after testing this method on a few other exercises I believe I understand it clearly.
            – Reiburn
            Nov 29 at 0:01




            Appreciate the response a lot, after testing this method on a few other exercises I believe I understand it clearly.
            – Reiburn
            Nov 29 at 0:01












            @Reiburn Glad to be of service!
            – JDMan4444
            Nov 29 at 0:03




            @Reiburn Glad to be of service!
            – JDMan4444
            Nov 29 at 0:03











            0














            You just apply basic power rules:

            $a^na^m = a^{n+m}$ etc ...


            Let's apply them now to your example:
            $$8a^2b^3(27)a^2b^3a^4ab$$
            $$8 (27) 2^5 a^2 b^3 a^4 ab$$
            $$2^3 2^5 (27) a^7 b^4$$
            $$2^8 3^3 a^7 b^4$$






            share|cite|improve this answer


























              0














              You just apply basic power rules:

              $a^na^m = a^{n+m}$ etc ...


              Let's apply them now to your example:
              $$8a^2b^3(27)a^2b^3a^4ab$$
              $$8 (27) 2^5 a^2 b^3 a^4 ab$$
              $$2^3 2^5 (27) a^7 b^4$$
              $$2^8 3^3 a^7 b^4$$






              share|cite|improve this answer
























                0












                0








                0






                You just apply basic power rules:

                $a^na^m = a^{n+m}$ etc ...


                Let's apply them now to your example:
                $$8a^2b^3(27)a^2b^3a^4ab$$
                $$8 (27) 2^5 a^2 b^3 a^4 ab$$
                $$2^3 2^5 (27) a^7 b^4$$
                $$2^8 3^3 a^7 b^4$$






                share|cite|improve this answer












                You just apply basic power rules:

                $a^na^m = a^{n+m}$ etc ...


                Let's apply them now to your example:
                $$8a^2b^3(27)a^2b^3a^4ab$$
                $$8 (27) 2^5 a^2 b^3 a^4 ab$$
                $$2^3 2^5 (27) a^7 b^4$$
                $$2^8 3^3 a^7 b^4$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 at 0:01









                Blg Khalil

                283




                283























                    0














                    I also have another way to do it. If you distribute the $8 a^2 b^3$ with the $27 a^4$, you will get $216 a^6 b^3$. Then, distribute that value with $2^5 a b$ and you will get $6,912 a^7 b^4$. The number $6,912$ can be written exponentially as $(2^8)(3^3)$. Bring that value down with the $a^7 b^4$, and you get your answer.






                    share|cite|improve this answer



















                    • 4




                      "Distribute"? In my experience, that word usually means something quite different from what you're doing here.
                      – David K
                      Nov 29 at 0:08












                    • By distribute, I really just mean multiply. When you do normal multiplication, you can really think of it is distributing between monomials, which is why it sounds different.
                      – Xavier Stanton
                      Nov 29 at 0:17






                    • 1




                      I got what you meant; the benefit of "thinking of" definitions in this way just isn't clear to me.
                      – David K
                      Nov 29 at 0:21
















                    0














                    I also have another way to do it. If you distribute the $8 a^2 b^3$ with the $27 a^4$, you will get $216 a^6 b^3$. Then, distribute that value with $2^5 a b$ and you will get $6,912 a^7 b^4$. The number $6,912$ can be written exponentially as $(2^8)(3^3)$. Bring that value down with the $a^7 b^4$, and you get your answer.






                    share|cite|improve this answer



















                    • 4




                      "Distribute"? In my experience, that word usually means something quite different from what you're doing here.
                      – David K
                      Nov 29 at 0:08












                    • By distribute, I really just mean multiply. When you do normal multiplication, you can really think of it is distributing between monomials, which is why it sounds different.
                      – Xavier Stanton
                      Nov 29 at 0:17






                    • 1




                      I got what you meant; the benefit of "thinking of" definitions in this way just isn't clear to me.
                      – David K
                      Nov 29 at 0:21














                    0












                    0








                    0






                    I also have another way to do it. If you distribute the $8 a^2 b^3$ with the $27 a^4$, you will get $216 a^6 b^3$. Then, distribute that value with $2^5 a b$ and you will get $6,912 a^7 b^4$. The number $6,912$ can be written exponentially as $(2^8)(3^3)$. Bring that value down with the $a^7 b^4$, and you get your answer.






                    share|cite|improve this answer














                    I also have another way to do it. If you distribute the $8 a^2 b^3$ with the $27 a^4$, you will get $216 a^6 b^3$. Then, distribute that value with $2^5 a b$ and you will get $6,912 a^7 b^4$. The number $6,912$ can be written exponentially as $(2^8)(3^3)$. Bring that value down with the $a^7 b^4$, and you get your answer.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 29 at 2:38









                    timtfj

                    977317




                    977317










                    answered Nov 29 at 0:04









                    Xavier Stanton

                    330211




                    330211








                    • 4




                      "Distribute"? In my experience, that word usually means something quite different from what you're doing here.
                      – David K
                      Nov 29 at 0:08












                    • By distribute, I really just mean multiply. When you do normal multiplication, you can really think of it is distributing between monomials, which is why it sounds different.
                      – Xavier Stanton
                      Nov 29 at 0:17






                    • 1




                      I got what you meant; the benefit of "thinking of" definitions in this way just isn't clear to me.
                      – David K
                      Nov 29 at 0:21














                    • 4




                      "Distribute"? In my experience, that word usually means something quite different from what you're doing here.
                      – David K
                      Nov 29 at 0:08












                    • By distribute, I really just mean multiply. When you do normal multiplication, you can really think of it is distributing between monomials, which is why it sounds different.
                      – Xavier Stanton
                      Nov 29 at 0:17






                    • 1




                      I got what you meant; the benefit of "thinking of" definitions in this way just isn't clear to me.
                      – David K
                      Nov 29 at 0:21








                    4




                    4




                    "Distribute"? In my experience, that word usually means something quite different from what you're doing here.
                    – David K
                    Nov 29 at 0:08






                    "Distribute"? In my experience, that word usually means something quite different from what you're doing here.
                    – David K
                    Nov 29 at 0:08














                    By distribute, I really just mean multiply. When you do normal multiplication, you can really think of it is distributing between monomials, which is why it sounds different.
                    – Xavier Stanton
                    Nov 29 at 0:17




                    By distribute, I really just mean multiply. When you do normal multiplication, you can really think of it is distributing between monomials, which is why it sounds different.
                    – Xavier Stanton
                    Nov 29 at 0:17




                    1




                    1




                    I got what you meant; the benefit of "thinking of" definitions in this way just isn't clear to me.
                    – David K
                    Nov 29 at 0:21




                    I got what you meant; the benefit of "thinking of" definitions in this way just isn't clear to me.
                    – David K
                    Nov 29 at 0:21


















                    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%2f3017928%2fbasic-mathematics-rules-for-multiplication-trouble%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

                    To store a contact into the json file from server.js file using a class in NodeJS

                    Image
                    0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
                    Read more

                    Redirect URL with Chrome Remote Debugging Android Devices

                    Image
                    1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
                    Read more

                    Dieringhausen

                    Image
                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                    Read more