Addition and subtraction with exponents
$begingroup$
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
asked Nov 28 '13 at 20:32
ThreethumbThreethumb
43631631
$endgroup$
add a comment |
$begingroup$
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
asked Nov 28 '13 at 20:32
ThreethumbThreethumb
43631631
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Nov 28 '13 at 20:32
ThreethumbThreethumb
43631631
$endgroup$
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
asked Nov 28 '13 at 20:32
ThreethumbThreethumb
43631631
asked Nov 28 '13 at 20:32
ThreethumbThreethumb
43631631
asked Nov 28 '13 at 20:32
ThreethumbThreethumb
43631631
asked Nov 28 '13 at 20:32
ThreethumbThreethumb
43631631
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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:
$$a^{m+1}-a^{m}=a^m(a-1).$$
answered Nov 28 '13 at 20:37
JP McCarthyJP McCarthy
5,73412441
$endgroup$
$begingroup$
Ah, yes, that makes sense. I get, this'll help! Thanks!
$endgroup$
– Threethumb
Nov 28 '13 at 21:23
add a comment |
$begingroup$
Yes there is:
$$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$
answered Nov 28 '13 at 20:40
LeeNeverGupLeeNeverGup
2,146717
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
$$a^{m+1}-a^{m}=a^m(a-1).$$
answered Nov 28 '13 at 20:37
JP McCarthyJP McCarthy
5,73412441
$endgroup$
$begingroup$
Ah, yes, that makes sense. I get, this'll help! Thanks!
$endgroup$
– Threethumb
Nov 28 '13 at 21:23
add a comment |
$begingroup$
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:
$$a^{m+1}-a^{m}=a^m(a-1).$$
answered Nov 28 '13 at 20:37
JP McCarthyJP McCarthy
5,73412441
$endgroup$
$begingroup$
Ah, yes, that makes sense. I get, this'll help! Thanks!
$endgroup$
– Threethumb
Nov 28 '13 at 21:23
add a comment |
$begingroup$
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:
$$a^{m+1}-a^{m}=a^m(a-1).$$
answered Nov 28 '13 at 20:37
JP McCarthyJP McCarthy
5,73412441
$endgroup$
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:
$$a^{m+1}-a^{m}=a^m(a-1).$$
answered Nov 28 '13 at 20:37
JP McCarthyJP McCarthy
5,73412441
answered Nov 28 '13 at 20:37
JP McCarthyJP McCarthy
5,73412441
answered Nov 28 '13 at 20:37
JP McCarthyJP McCarthy
5,73412441
answered Nov 28 '13 at 20:37
JP McCarthyJP McCarthy
5,73412441
5,73412441
$begingroup$
Ah, yes, that makes sense. I get, this'll help! Thanks!
$endgroup$
– Threethumb
Nov 28 '13 at 21:23
add a comment |
$begingroup$
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
add a comment |
$begingroup$
Yes there is:
$$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$
answered Nov 28 '13 at 20:40
LeeNeverGupLeeNeverGup
2,146717
$endgroup$
add a comment |
$begingroup$
Yes there is:
$$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$
answered Nov 28 '13 at 20:40
LeeNeverGupLeeNeverGup
2,146717
$endgroup$
add a comment |
$begingroup$
Yes there is:
$$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$
answered Nov 28 '13 at 20:40
LeeNeverGupLeeNeverGup
2,146717
$endgroup$
Yes there is:
$$2^7-2^6=2*2^6-2^6=(2-1)2^6=2^6$$
answered Nov 28 '13 at 20:40
LeeNeverGupLeeNeverGup
2,146717
answered Nov 28 '13 at 20:40
LeeNeverGupLeeNeverGup
2,146717
answered Nov 28 '13 at 20:40
LeeNeverGupLeeNeverGup
2,146717
answered Nov 28 '13 at 20:40
LeeNeverGupLeeNeverGup
2,146717
2,146717
add a comment |
add a comment |
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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