Power on absolute value expressions
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How to deal with absolute value, when raised to odd number?
Like here:
$$left|log_2left(frac{x}{2}right)right|^3 + |log_2 (2x)|^3 = 28$$
algebra-precalculus absolute-value
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up vote
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How to deal with absolute value, when raised to odd number?
Like here:
$$left|log_2left(frac{x}{2}right)right|^3 + |log_2 (2x)|^3 = 28$$
algebra-precalculus absolute-value
New contributor
Well, as you know, odd powers preserve the sign of the thing that you're raising to a power. So ... that's it. You need to keep the absolute values there and consider the different cases as usual.
– Matti P.
yesterday
Please include the associated problem in the question body. Links have a tendency to die and make for poor questions for someone coming back years from now.
– DreamConspiracy
yesterday
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr Samir
yesterday
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
yesterday
Thanks alot. @N.F.Taussig
– Amr Samir
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to deal with absolute value, when raised to odd number?
Like here:
$$left|log_2left(frac{x}{2}right)right|^3 + |log_2 (2x)|^3 = 28$$
algebra-precalculus absolute-value
New contributor
How to deal with absolute value, when raised to odd number?
Like here:
$$left|log_2left(frac{x}{2}right)right|^3 + |log_2 (2x)|^3 = 28$$
algebra-precalculus absolute-value
algebra-precalculus absolute-value
New contributor
New contributor
edited yesterday
N. F. Taussig
42.6k93254
42.6k93254
New contributor
asked yesterday
Amr Samir
101
101
New contributor
New contributor
Well, as you know, odd powers preserve the sign of the thing that you're raising to a power. So ... that's it. You need to keep the absolute values there and consider the different cases as usual.
– Matti P.
yesterday
Please include the associated problem in the question body. Links have a tendency to die and make for poor questions for someone coming back years from now.
– DreamConspiracy
yesterday
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr Samir
yesterday
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
yesterday
Thanks alot. @N.F.Taussig
– Amr Samir
yesterday
add a comment |
Well, as you know, odd powers preserve the sign of the thing that you're raising to a power. So ... that's it. You need to keep the absolute values there and consider the different cases as usual.
– Matti P.
yesterday
Please include the associated problem in the question body. Links have a tendency to die and make for poor questions for someone coming back years from now.
– DreamConspiracy
yesterday
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr Samir
yesterday
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
yesterday
Thanks alot. @N.F.Taussig
– Amr Samir
yesterday
Well, as you know, odd powers preserve the sign of the thing that you're raising to a power. So ... that's it. You need to keep the absolute values there and consider the different cases as usual.
– Matti P.
yesterday
Well, as you know, odd powers preserve the sign of the thing that you're raising to a power. So ... that's it. You need to keep the absolute values there and consider the different cases as usual.
– Matti P.
yesterday
Please include the associated problem in the question body. Links have a tendency to die and make for poor questions for someone coming back years from now.
– DreamConspiracy
yesterday
Please include the associated problem in the question body. Links have a tendency to die and make for poor questions for someone coming back years from now.
– DreamConspiracy
yesterday
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr Samir
yesterday
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr Samir
yesterday
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
yesterday
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
yesterday
Thanks alot. @N.F.Taussig
– Amr Samir
yesterday
Thanks alot. @N.F.Taussig
– Amr Samir
yesterday
add a comment |
1 Answer
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accepted
We have $log_2left(frac{x}{2}right)=log_2 x-1; text{and} ; log_2left(2xright)=log_2 x+1.$
Set $a=log_2x.$ The equation rewrites
$$left|a-1right|^3 + |a+1|^3 = 28.$$
Solve separately for $; a<-1,; -1leq a leq 1, ; a>1.$ We get $a=-2$ or $a=2.$
The solutions of the initial equation $xin{{1over 4}, 4}.$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have $log_2left(frac{x}{2}right)=log_2 x-1; text{and} ; log_2left(2xright)=log_2 x+1.$
Set $a=log_2x.$ The equation rewrites
$$left|a-1right|^3 + |a+1|^3 = 28.$$
Solve separately for $; a<-1,; -1leq a leq 1, ; a>1.$ We get $a=-2$ or $a=2.$
The solutions of the initial equation $xin{{1over 4}, 4}.$
add a comment |
up vote
1
down vote
accepted
We have $log_2left(frac{x}{2}right)=log_2 x-1; text{and} ; log_2left(2xright)=log_2 x+1.$
Set $a=log_2x.$ The equation rewrites
$$left|a-1right|^3 + |a+1|^3 = 28.$$
Solve separately for $; a<-1,; -1leq a leq 1, ; a>1.$ We get $a=-2$ or $a=2.$
The solutions of the initial equation $xin{{1over 4}, 4}.$
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have $log_2left(frac{x}{2}right)=log_2 x-1; text{and} ; log_2left(2xright)=log_2 x+1.$
Set $a=log_2x.$ The equation rewrites
$$left|a-1right|^3 + |a+1|^3 = 28.$$
Solve separately for $; a<-1,; -1leq a leq 1, ; a>1.$ We get $a=-2$ or $a=2.$
The solutions of the initial equation $xin{{1over 4}, 4}.$
We have $log_2left(frac{x}{2}right)=log_2 x-1; text{and} ; log_2left(2xright)=log_2 x+1.$
Set $a=log_2x.$ The equation rewrites
$$left|a-1right|^3 + |a+1|^3 = 28.$$
Solve separately for $; a<-1,; -1leq a leq 1, ; a>1.$ We get $a=-2$ or $a=2.$
The solutions of the initial equation $xin{{1over 4}, 4}.$
answered yesterday
user376343
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2,2531716
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add a comment |
Amr Samir is a new contributor. Be nice, and check out our Code of Conduct.
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Well, as you know, odd powers preserve the sign of the thing that you're raising to a power. So ... that's it. You need to keep the absolute values there and consider the different cases as usual.
– Matti P.
yesterday
Please include the associated problem in the question body. Links have a tendency to die and make for poor questions for someone coming back years from now.
– DreamConspiracy
yesterday
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr Samir
yesterday
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
yesterday
Thanks alot. @N.F.Taussig
– Amr Samir
yesterday