Power on absolute value expressions











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How to deal with absolute value, when raised to odd number?



Like here:
$$left|log_2left(frac{x}{2}right)right|^3 + |log_2 (2x)|^3 = 28$$










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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 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 at 9:12










  • @DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
    – Amr S. Omar
    Nov 21 at 11:51










  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 21 at 14:28










  • Thanks alot. @N.F.Taussig
    – Amr S. Omar
    Nov 21 at 14:30















up vote
0
down vote

favorite












How to deal with absolute value, when raised to odd number?



Like here:
$$left|log_2left(frac{x}{2}right)right|^3 + |log_2 (2x)|^3 = 28$$










share|cite|improve this question
























  • 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 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 at 9:12










  • @DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
    – Amr S. Omar
    Nov 21 at 11:51










  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 21 at 14:28










  • Thanks alot. @N.F.Taussig
    – Amr S. Omar
    Nov 21 at 14:30













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:
$$left|log_2left(frac{x}{2}right)right|^3 + |log_2 (2x)|^3 = 28$$










share|cite|improve this question















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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share|cite|improve this question













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edited Nov 21 at 14:27









N. F. Taussig

43.1k93254




43.1k93254










asked Nov 21 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 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 at 9:12










  • @DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
    – Amr S. Omar
    Nov 21 at 11:51










  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 21 at 14:28










  • Thanks alot. @N.F.Taussig
    – Amr S. Omar
    Nov 21 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 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 at 9:12










  • @DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
    – Amr S. Omar
    Nov 21 at 11:51










  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 21 at 14:28










  • Thanks alot. @N.F.Taussig
    – Amr S. Omar
    Nov 21 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 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 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 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 at 9:12












@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr S. Omar
Nov 21 at 11:51




@DreamConspiracy Sorry, I'm having a problem formatting the equation . You may edit the question .
– Amr S. Omar
Nov 21 at 11:51












Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 21 at 14:28




Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 21 at 14:28












Thanks alot. @N.F.Taussig
– Amr S. Omar
Nov 21 at 14:30




Thanks alot. @N.F.Taussig
– Amr S. Omar
Nov 21 at 14:30










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 $xin{{1over 4}, 4}.$






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    1 Answer
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    1 Answer
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    votes






    active

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    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 $xin{{1over 4}, 4}.$






    share|cite|improve this answer

























      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 $xin{{1over 4}, 4}.$






      share|cite|improve this answer























        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 $xin{{1over 4}, 4}.$






        share|cite|improve this answer












        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 $xin{{1over 4}, 4}.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 20:56









        user376343

        2,6512819




        2,6512819






























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