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





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            3 Answers
            3






            active

            oldest

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            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


















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