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.
    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















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









New contributor




Amr Samir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 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













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









New contributor




Amr Samir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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






share|cite|improve this question









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Amr Samir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Amr Samir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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edited yesterday









N. F. Taussig

42.6k93254




42.6k93254






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asked yesterday









Amr Samir

101




101




New contributor




Amr Samir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Amr Samir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Amr Samir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • 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










  • 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










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






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    1 Answer
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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 yesterday









        user376343

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