How do you simplify a fraction raised to a negative exponent?
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$left( frac23 right) ^-2$ I noticed that the answer is $frac94$, and I can't come to the conclusion of why it is?
arithmetic
edited Sep 15 '17 at 19:57
JMoravitz
44.5k33481
asked Sep 15 '17 at 19:49
Jorden Danczak
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$left( frac23 right) ^-2$ I noticed that the answer is $frac94$, and I can't come to the conclusion of why it is?
arithmetic
edited Sep 15 '17 at 19:57
JMoravitz
44.5k33481
asked Sep 15 '17 at 19:49
Jorden Danczak
1
The verb "solve" is inappropriate here. We are not solving anything. We are evaluating or simplifying or expressing differently.
– JMoravitz
Sep 15 '17 at 19:55
You need to keep in mind the following : $$forall x,yinmathbbR, x^-y=1over x^y$$ Now you can substitute $x$ and $y$ with your values and you'll find the answer =)
– Furrane
Sep 15 '17 at 19:59
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$left( frac23 right) ^-2$ I noticed that the answer is $frac94$, and I can't come to the conclusion of why it is?
arithmetic
edited Sep 15 '17 at 19:57
JMoravitz
44.5k33481
asked Sep 15 '17 at 19:49
Jorden Danczak
1
$left( frac23 right) ^-2$ I noticed that the answer is $frac94$, and I can't come to the conclusion of why it is?
arithmetic
edited Sep 15 '17 at 19:57
JMoravitz
44.5k33481
asked Sep 15 '17 at 19:49
Jorden Danczak
1
edited Sep 15 '17 at 19:57
JMoravitz
44.5k33481
edited Sep 15 '17 at 19:57
JMoravitz
44.5k33481
edited Sep 15 '17 at 19:57
JMoravitz
44.5k33481
44.5k33481
asked Sep 15 '17 at 19:49
Jorden Danczak
1
asked Sep 15 '17 at 19:49
Jorden Danczak
1
asked Sep 15 '17 at 19:49
Jorden Danczak
1
1
The verb "solve" is inappropriate here. We are not solving anything. We are evaluating or simplifying or expressing differently.
– JMoravitz
Sep 15 '17 at 19:55
You need to keep in mind the following : $$forall x,yinmathbbR, x^-y=1over x^y$$ Now you can substitute $x$ and $y$ with your values and you'll find the answer =)
– Furrane
Sep 15 '17 at 19:59
add a comment |Â
The verb "solve" is inappropriate here. We are not solving anything. We are evaluating or simplifying or expressing differently.
– JMoravitz
Sep 15 '17 at 19:55
You need to keep in mind the following : $$forall x,yinmathbbR, x^-y=1over x^y$$ Now you can substitute $x$ and $y$ with your values and you'll find the answer =)
– Furrane
Sep 15 '17 at 19:59
The verb "solve" is inappropriate here. We are not solving anything. We are evaluating or simplifying or expressing differently.
– JMoravitz
Sep 15 '17 at 19:55
The verb "solve" is inappropriate here. We are not solving anything. We are evaluating or simplifying or expressing differently.
– JMoravitz
Sep 15 '17 at 19:55
You need to keep in mind the following : $$forall x,yinmathbbR, x^-y=1over x^y$$ Now you can substitute $x$ and $y$ with your values and you'll find the answer =)
– Furrane
Sep 15 '17 at 19:59
You need to keep in mind the following : $$forall x,yinmathbbR, x^-y=1over x^y$$ Now you can substitute $x$ and $y$ with your values and you'll find the answer =)
– Furrane
Sep 15 '17 at 19:59
add a comment |Â
3 Answers
3
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oldest
votes
up vote
1
down vote
Guide:
Approach $1$:
Well, first evaluate $left( frac23right)^-1= fracab$
After which you can compute $left(fracabright)^2=fraca^2b^2$
Approach $2$:
Well, first evaluate $left( frac23right)^2= fraccd$
After which you can compute $left(fraccdright)^-1=fracdc$
answered Sep 15 '17 at 19:55
Siong Thye Goh
79.1k134997
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$$x^-1=frac 1x$$
Thus $$left(frac 23 right)^-2=left(left(frac 23right)^2right)^-1=left(frac 49 right)^-1=frac 94$$
answered Sep 15 '17 at 19:54
Jaideep Khare
17.6k32265
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Because $;displaystyleBigl(fracabBigr)^-2=biggl(Bigl(fracabBigr)^-1biggr)^2=Bigl(fracbaBigr)^2=fracb^2a^2.$
answered Sep 15 '17 at 19:56
Bernard
111k635103
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Guide:
Approach $1$:
Well, first evaluate $left( frac23right)^-1= fracab$
After which you can compute $left(fracabright)^2=fraca^2b^2$
Approach $2$:
Well, first evaluate $left( frac23right)^2= fraccd$
After which you can compute $left(fraccdright)^-1=fracdc$
answered Sep 15 '17 at 19:55
Siong Thye Goh
79.1k134997
add a comment |Â
up vote
1
down vote
Guide:
Approach $1$:
Well, first evaluate $left( frac23right)^-1= fracab$
After which you can compute $left(fracabright)^2=fraca^2b^2$
Approach $2$:
Well, first evaluate $left( frac23right)^2= fraccd$
After which you can compute $left(fraccdright)^-1=fracdc$
answered Sep 15 '17 at 19:55
Siong Thye Goh
79.1k134997
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Guide:
Approach $1$:
Well, first evaluate $left( frac23right)^-1= fracab$
After which you can compute $left(fracabright)^2=fraca^2b^2$
Approach $2$:
Well, first evaluate $left( frac23right)^2= fraccd$
After which you can compute $left(fraccdright)^-1=fracdc$
answered Sep 15 '17 at 19:55
Siong Thye Goh
79.1k134997
Guide:
Approach $1$:
Well, first evaluate $left( frac23right)^-1= fracab$
After which you can compute $left(fracabright)^2=fraca^2b^2$
Approach $2$:
Well, first evaluate $left( frac23right)^2= fraccd$
After which you can compute $left(fraccdright)^-1=fracdc$
answered Sep 15 '17 at 19:55
Siong Thye Goh
79.1k134997
answered Sep 15 '17 at 19:55
Siong Thye Goh
79.1k134997
answered Sep 15 '17 at 19:55
Siong Thye Goh
79.1k134997
answered Sep 15 '17 at 19:55
Siong Thye Goh
79.1k134997
79.1k134997
add a comment |Â
add a comment |Â
up vote
0
down vote
$$x^-1=frac 1x$$
Thus $$left(frac 23 right)^-2=left(left(frac 23right)^2right)^-1=left(frac 49 right)^-1=frac 94$$
answered Sep 15 '17 at 19:54
Jaideep Khare
17.6k32265
add a comment |Â
up vote
0
down vote
$$x^-1=frac 1x$$
Thus $$left(frac 23 right)^-2=left(left(frac 23right)^2right)^-1=left(frac 49 right)^-1=frac 94$$
answered Sep 15 '17 at 19:54
Jaideep Khare
17.6k32265
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$x^-1=frac 1x$$
Thus $$left(frac 23 right)^-2=left(left(frac 23right)^2right)^-1=left(frac 49 right)^-1=frac 94$$
answered Sep 15 '17 at 19:54
Jaideep Khare
17.6k32265
$$x^-1=frac 1x$$
Thus $$left(frac 23 right)^-2=left(left(frac 23right)^2right)^-1=left(frac 49 right)^-1=frac 94$$
answered Sep 15 '17 at 19:54
Jaideep Khare
17.6k32265
answered Sep 15 '17 at 19:54
Jaideep Khare
17.6k32265
answered Sep 15 '17 at 19:54
Jaideep Khare
17.6k32265
answered Sep 15 '17 at 19:54
Jaideep Khare
17.6k32265
17.6k32265
add a comment |Â
add a comment |Â
up vote
0
down vote
Because $;displaystyleBigl(fracabBigr)^-2=biggl(Bigl(fracabBigr)^-1biggr)^2=Bigl(fracbaBigr)^2=fracb^2a^2.$
answered Sep 15 '17 at 19:56
Bernard
111k635103
add a comment |Â
up vote
0
down vote
Because $;displaystyleBigl(fracabBigr)^-2=biggl(Bigl(fracabBigr)^-1biggr)^2=Bigl(fracbaBigr)^2=fracb^2a^2.$
answered Sep 15 '17 at 19:56
Bernard
111k635103
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Because $;displaystyleBigl(fracabBigr)^-2=biggl(Bigl(fracabBigr)^-1biggr)^2=Bigl(fracbaBigr)^2=fracb^2a^2.$
answered Sep 15 '17 at 19:56
Bernard
111k635103
Because $;displaystyleBigl(fracabBigr)^-2=biggl(Bigl(fracabBigr)^-1biggr)^2=Bigl(fracbaBigr)^2=fracb^2a^2.$
answered Sep 15 '17 at 19:56
Bernard
111k635103
answered Sep 15 '17 at 19:56
Bernard
111k635103
answered Sep 15 '17 at 19:56
Bernard
111k635103
answered Sep 15 '17 at 19:56
Bernard
111k635103
111k635103
add a comment |Â
add a comment |Â
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The verb "solve" is inappropriate here. We are not solving anything. We are evaluating or simplifying or expressing differently.
– JMoravitz
Sep 15 '17 at 19:55
You need to keep in mind the following : $$forall x,yinmathbbR, x^-y=1over x^y$$ Now you can substitute $x$ and $y$ with your values and you'll find the answer =)
– Furrane
Sep 15 '17 at 19:59