Justification of exponent laws
Clash Royale CLAN TAG#URR8PPP
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How is the following operation justified? Specifically, what happens to the $4^n$ in the numerator. I tried looking online but could not determine how this was justified. Thanks.
$$dfrac(-3)^n-14^n= dfrac14cdot left(-dfrac34right)^n-1$$
exponential-function
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up vote
3
down vote
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How is the following operation justified? Specifically, what happens to the $4^n$ in the numerator. I tried looking online but could not determine how this was justified. Thanks.
$$dfrac(-3)^n-14^n= dfrac14cdot left(-dfrac34right)^n-1$$
exponential-function
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How is the following operation justified? Specifically, what happens to the $4^n$ in the numerator. I tried looking online but could not determine how this was justified. Thanks.
$$dfrac(-3)^n-14^n= dfrac14cdot left(-dfrac34right)^n-1$$
exponential-function
How is the following operation justified? Specifically, what happens to the $4^n$ in the numerator. I tried looking online but could not determine how this was justified. Thanks.
$$dfrac(-3)^n-14^n= dfrac14cdot left(-dfrac34right)^n-1$$
exponential-function
exponential-function
edited Apr 5 '15 at 22:02
N. F. Taussig
39.9k93253
39.9k93253
asked Apr 5 '15 at 20:26
user3826764
182
182
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1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
The $4^n$ is in the denominator.
It becomes $4cdot 4^n-1$
The $4^n-1$ is then combined with the $(-3)^n-1$ in the numerator.
@N.F.Taussig: thank you for the $$ $
â Henry
Apr 5 '15 at 23:42
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The $4^n$ is in the denominator.
It becomes $4cdot 4^n-1$
The $4^n-1$ is then combined with the $(-3)^n-1$ in the numerator.
@N.F.Taussig: thank you for the $$ $
â Henry
Apr 5 '15 at 23:42
add a comment |Â
up vote
0
down vote
accepted
The $4^n$ is in the denominator.
It becomes $4cdot 4^n-1$
The $4^n-1$ is then combined with the $(-3)^n-1$ in the numerator.
@N.F.Taussig: thank you for the $$ $
â Henry
Apr 5 '15 at 23:42
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The $4^n$ is in the denominator.
It becomes $4cdot 4^n-1$
The $4^n-1$ is then combined with the $(-3)^n-1$ in the numerator.
The $4^n$ is in the denominator.
It becomes $4cdot 4^n-1$
The $4^n-1$ is then combined with the $(-3)^n-1$ in the numerator.
edited Apr 5 '15 at 22:03
N. F. Taussig
39.9k93253
39.9k93253
answered Apr 5 '15 at 20:29
Henry
94.5k473150
94.5k473150
@N.F.Taussig: thank you for the $$ $
â Henry
Apr 5 '15 at 23:42
add a comment |Â
@N.F.Taussig: thank you for the $$ $
â Henry
Apr 5 '15 at 23:42
@N.F.Taussig: thank you for the $$ $
â Henry
Apr 5 '15 at 23:42
@N.F.Taussig: thank you for the $$ $
â Henry
Apr 5 '15 at 23:42
add a comment |Â
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