Addition and subtraction with exponents












1












$begingroup$


I'm doing an Advanced Functions course right now, and I'm wondering about something. Look at this here evaluation/simplification that I did:



http://puu.sh/5w3XQ.png



What I'm wondering is about the 2^4 - 2^3. I know this is is 2^3 because 2^3 is the value multiplied by 2 to get 2^4, which means I'm subtracting the same value I'm adding when going from 2^3 to 2^4 from 2^3 and thereby going back to 2^3.



But this is just a logical step I did it my head, and it's not something that works as a rule for all addition/subtraction. So I'm wondering if there is another way for me to solve this? I'm not comfortable with simplifying something on the premise of "I just know that's what it means", because that might mean I don't get the full score. So is there any exponent rules/laws that apply to adding or subtracting exponents with the same base which I can use here, in order to make sure I don't get less point for taking shortcuts instead of using the exponent laws?



Should I perhaps make it into 2^7 - 2^6 and evaluate it as 128 - 64 = 64? Or is there another way to end up with 2^6 using an exponent law? I just want to make sure I don't get any less points for doing steps in my head or simplifying something outside of the methods of exponent laws.






arithmetic







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  • $begingroup$
    What you did looks correct, but..."Advanced" functions??
    $endgroup$
    – DonAntonio
    Nov 28 '13 at 20:35










  • $begingroup$
    That's what the course is called, but I'm only at the beginning which is mostly focused around re-iterating the basics.
    $endgroup$
    – Threethumb
    Nov 28 '13 at 21:19
















1












$begingroup$


I'm doing an Advanced Functions course right now, and I'm wondering about something. Look at this here evaluation/simplification that I did:



http://puu.sh/5w3XQ.png



What I'm wondering is about the 2^4 - 2^3. I know this is is 2^3 because 2^3 is the value multiplied by 2 to get 2^4, which means I'm subtracting the same value I'm adding when going from 2^3 to 2^4 from 2^3 and thereby going back to 2^3.



But this is just a logical step I did it my head, and it's not something that works as a rule for all addition/subtraction. So I'm wondering if there is another way for me to solve this? I'm not comfortable with simplifying something on the premise of "I just know that's what it means", because that might mean I don't get the full score. So is there any exponent rules/laws that apply to adding or subtracting exponents with the same base which I can use here, in order to make sure I don't get less point for taking shortcuts instead of using the exponent laws?



Should I perhaps make it into 2^7 - 2^6 and evaluate it as 128 - 64 = 64? Or is there another way to end up with 2^6 using an exponent law? I just want to make sure I don't get any less points for doing steps in my head or simplifying something outside of the methods of exponent laws.






arithmetic







share|cite|improve this question









$endgroup$












  • $begingroup$
    What you did looks correct, but..."Advanced" functions??
    $endgroup$
    – DonAntonio
    Nov 28 '13 at 20:35










  • $begingroup$
    That's what the course is called, but I'm only at the beginning which is mostly focused around re-iterating the basics.
    $endgroup$
    – Threethumb
    Nov 28 '13 at 21:19














1












1








1





$begingroup$


I'm doing an Advanced Functions course right now, and I'm wondering about something. Look at this here evaluation/simplification that I did:



http://puu.sh/5w3XQ.png



What I'm wondering is about the 2^4 - 2^3. I know this is is 2^3 because 2^3 is the value multiplied by 2 to get 2^4, which means I'm subtracting the same value I'm adding when going from 2^3 to 2^4 from 2^3 and thereby going back to 2^3.



But this is just a logical step I did it my head, and it's not something that works as a rule for all addition/subtraction. So I'm wondering if there is another way for me to solve this? I'm not comfortable with simplifying something on the premise of "I just know that's what it means", because that might mean I don't get the full score. So is there any exponent rules/laws that apply to adding or subtracting exponents with the same base which I can use here, in order to make sure I don't get less point for taking shortcuts instead of using the exponent laws?



Should I perhaps make it into 2^7 - 2^6 and evaluate it as 128 - 64 = 64? Or is there another way to end up with 2^6 using an exponent law? I just want to make sure I don't get any less points for doing steps in my head or simplifying something outside of the methods of exponent laws.






arithmetic







share|cite|improve this question









$endgroup$




I'm doing an Advanced Functions course right now, and I'm wondering about something. Look at this here evaluation/simplification that I did:



http://puu.sh/5w3XQ.png



What I'm wondering is about the 2^4 - 2^3. I know this is is 2^3 because 2^3 is the value multiplied by 2 to get 2^4, which means I'm subtracting the same value I'm adding when going from 2^3 to 2^4 from 2^3 and thereby going back to 2^3.



But this is just a logical step I did it my head, and it's not something that works as a rule for all addition/subtraction. So I'm wondering if there is another way for me to solve this? I'm not comfortable with simplifying something on the premise of "I just know that's what it means", because that might mean I don't get the full score. So is there any exponent rules/laws that apply to adding or subtracting exponents with the same base which I can use here, in order to make sure I don't get less point for taking shortcuts instead of using the exponent laws?



Should I perhaps make it into 2^7 - 2^6 and evaluate it as 128 - 64 = 64? Or is there another way to end up with 2^6 using an exponent law? I just want to make sure I don't get any less points for doing steps in my head or simplifying something outside of the methods of exponent laws.






arithmetic




arithmetic






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asked Nov 28 '13 at 20:32









ThreethumbThreethumb

43631631




43631631












  • $begingroup$
    What you did looks correct, but..."Advanced" functions??
    $endgroup$
    – DonAntonio
    Nov 28 '13 at 20:35










  • $begingroup$
    That's what the course is called, but I'm only at the beginning which is mostly focused around re-iterating the basics.
    $endgroup$
    – Threethumb
    Nov 28 '13 at 21:19


















  • $begingroup$
    What you did looks correct, but..."Advanced" functions??
    $endgroup$
    – DonAntonio
    Nov 28 '13 at 20:35










  • $begingroup$
    That's what the course is called, but I'm only at the beginning which is mostly focused around re-iterating the basics.
    $endgroup$
    – Threethumb
    Nov 28 '13 at 21:19
















$begingroup$
What you did looks correct, but..."Advanced" functions??
$endgroup$
– DonAntonio
Nov 28 '13 at 20:35




$begingroup$
What you did looks correct, but..."Advanced" functions??
$endgroup$
– DonAntonio
Nov 28 '13 at 20:35












$begingroup$
That's what the course is called, but I'm only at the beginning which is mostly focused around re-iterating the basics.
$endgroup$
– Threethumb
Nov 28 '13 at 21:19




$begingroup$
That's what the course is called, but I'm only at the beginning which is mostly focused around re-iterating the basics.
$endgroup$
– Threethumb
Nov 28 '13 at 21:19










2 Answers
2






active

oldest

votes


















0












$begingroup$

All you can really say is this. Suppose that $m>n$ and we are looking at



$$a^m+a^n.$$
By definition $m=n+k$ so both have a common factor of $a^n$:



$$a^m+a^n=a^{n+k}+a^n=a^{n}(a^k+1).$$



In your example,



$$2^4-2^3=2^3(2-1)=2^3.$$



However to be honest... $2^4=16$ and $2^{3}=8$ --- it is as simple as that.



Or a bit more general but closer to your example:
$$a^{m+1}-a^{m}=a^m(a-1).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah, yes, that makes sense. I get, this'll help! Thanks!
    $endgroup$
    – Threethumb
    Nov 28 '13 at 21:23



















0












$begingroup$

Yes there is:
$$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    All you can really say is this. Suppose that $m>n$ and we are looking at



    $$a^m+a^n.$$
    By definition $m=n+k$ so both have a common factor of $a^n$:



    $$a^m+a^n=a^{n+k}+a^n=a^{n}(a^k+1).$$



    In your example,



    $$2^4-2^3=2^3(2-1)=2^3.$$



    However to be honest... $2^4=16$ and $2^{3}=8$ --- it is as simple as that.



    Or a bit more general but closer to your example:
    $$a^{m+1}-a^{m}=a^m(a-1).$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ah, yes, that makes sense. I get, this'll help! Thanks!
      $endgroup$
      – Threethumb
      Nov 28 '13 at 21:23
















    0












    $begingroup$

    All you can really say is this. Suppose that $m>n$ and we are looking at



    $$a^m+a^n.$$
    By definition $m=n+k$ so both have a common factor of $a^n$:



    $$a^m+a^n=a^{n+k}+a^n=a^{n}(a^k+1).$$



    In your example,



    $$2^4-2^3=2^3(2-1)=2^3.$$



    However to be honest... $2^4=16$ and $2^{3}=8$ --- it is as simple as that.



    Or a bit more general but closer to your example:
    $$a^{m+1}-a^{m}=a^m(a-1).$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ah, yes, that makes sense. I get, this'll help! Thanks!
      $endgroup$
      – Threethumb
      Nov 28 '13 at 21:23














    0












    0








    0





    $begingroup$

    All you can really say is this. Suppose that $m>n$ and we are looking at



    $$a^m+a^n.$$
    By definition $m=n+k$ so both have a common factor of $a^n$:



    $$a^m+a^n=a^{n+k}+a^n=a^{n}(a^k+1).$$



    In your example,



    $$2^4-2^3=2^3(2-1)=2^3.$$



    However to be honest... $2^4=16$ and $2^{3}=8$ --- it is as simple as that.



    Or a bit more general but closer to your example:
    $$a^{m+1}-a^{m}=a^m(a-1).$$






    share|cite|improve this answer









    $endgroup$



    All you can really say is this. Suppose that $m>n$ and we are looking at



    $$a^m+a^n.$$
    By definition $m=n+k$ so both have a common factor of $a^n$:



    $$a^m+a^n=a^{n+k}+a^n=a^{n}(a^k+1).$$



    In your example,



    $$2^4-2^3=2^3(2-1)=2^3.$$



    However to be honest... $2^4=16$ and $2^{3}=8$ --- it is as simple as that.



    Or a bit more general but closer to your example:
    $$a^{m+1}-a^{m}=a^m(a-1).$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 '13 at 20:37









    JP McCarthyJP McCarthy

    5,73412442




    5,73412442












    • $begingroup$
      Ah, yes, that makes sense. I get, this'll help! Thanks!
      $endgroup$
      – Threethumb
      Nov 28 '13 at 21:23


















    • $begingroup$
      Ah, yes, that makes sense. I get, this'll help! Thanks!
      $endgroup$
      – Threethumb
      Nov 28 '13 at 21:23
















    $begingroup$
    Ah, yes, that makes sense. I get, this'll help! Thanks!
    $endgroup$
    – Threethumb
    Nov 28 '13 at 21:23




    $begingroup$
    Ah, yes, that makes sense. I get, this'll help! Thanks!
    $endgroup$
    – Threethumb
    Nov 28 '13 at 21:23











    0












    $begingroup$

    Yes there is:
    $$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Yes there is:
      $$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Yes there is:
        $$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$






        share|cite|improve this answer









        $endgroup$



        Yes there is:
        $$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '13 at 20:40









        LeeNeverGupLeeNeverGup

        2,146717




        2,146717






























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