Mixing
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Power on absolute value expressions
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
edited Nov 21 '18 at 14:27
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 8:18
Amr S. Omar
101
add a comment |
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
edited Nov 21 '18 at 14:27
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 8:18
Amr S. Omar
101
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.
Nov 21 '18 at 8:22
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
Nov 21 '18 at 9:12
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr S. Omar
Nov 21 '18 at 11:51
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 21 '18 at 14:28
Thanks alot. @N.F.Taussig
– Amr S. Omar
Nov 21 '18 at 14:30
add a comment |
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
edited Nov 21 '18 at 14:27
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 8:18
Amr S. Omar
101
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
edited Nov 21 '18 at 14:27
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 8:18
Amr S. Omar
101
edited Nov 21 '18 at 14:27
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 8:18
Amr S. Omar
101
edited Nov 21 '18 at 14:27
N. F. Taussig
43.6k93355
edited Nov 21 '18 at 14:27
N. F. Taussig
43.6k93355
edited Nov 21 '18 at 14:27
N. F. Taussig
43.6k93355
43.6k93355
asked Nov 21 '18 at 8:18
Amr S. Omar
101
asked Nov 21 '18 at 8:18
Amr S. Omar
101
asked Nov 21 '18 at 8:18
Amr S. Omar
101
101
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.
Nov 21 '18 at 8:22
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
Nov 21 '18 at 9:12
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr S. Omar
Nov 21 '18 at 11:51
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 21 '18 at 14:28
Thanks alot. @N.F.Taussig
– Amr S. Omar
Nov 21 '18 at 14:30
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.
Nov 21 '18 at 8:22
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
Nov 21 '18 at 9:12
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr S. Omar
Nov 21 '18 at 11:51
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 21 '18 at 14:28
Thanks alot. @N.F.Taussig
– Amr S. Omar
Nov 21 '18 at 14:30
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.
Nov 21 '18 at 8:22
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.
Nov 21 '18 at 8:22
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
Nov 21 '18 at 9:12
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
Nov 21 '18 at 9:12
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr S. Omar
Nov 21 '18 at 11:51
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr S. Omar
Nov 21 '18 at 11:51
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 21 '18 at 14:28
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 21 '18 at 14:28
Thanks alot. @N.F.Taussig
– Amr S. Omar
Nov 21 '18 at 14:30
Thanks alot. @N.F.Taussig
– Amr S. Omar
Nov 21 '18 at 14:30
add a comment |
1 Answer
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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 Nov 21 '18 at 20:56
user376343
2,8582823
add a comment |
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1 Answer
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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 Nov 21 '18 at 20:56
user376343
2,8582823
add a comment |
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 Nov 21 '18 at 20:56
user376343
2,8582823
add a comment |
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 Nov 21 '18 at 20:56
user376343
2,8582823
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 Nov 21 '18 at 20:56
user376343
2,8582823
answered Nov 21 '18 at 20:56
user376343
2,8582823
answered Nov 21 '18 at 20:56
user376343
2,8582823
answered Nov 21 '18 at 20:56
user376343
2,8582823
2,8582823
add a comment |
add a comment |
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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.
Nov 21 '18 at 8:22
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
Nov 21 '18 at 9:12
@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr S. Omar
Nov 21 '18 at 11:51
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 21 '18 at 14:28
Thanks alot. @N.F.Taussig
– Amr S. Omar
Nov 21 '18 at 14:30