Is it the same $a^alpha + b^alpha = x$ as $(a^alpha)^frac1alpha + (b^alpha)^frac1alpha = x^frac1alpha$

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I want to ask if it is possible to transform equation $a^alpha + b^alpha = x$ into $a + b = x^frac1alpha$ by elevating each parameter by $frac1alpha$? Or if we elevate each parameter by $frac1alpha$ it must be $(a^alpha + b^alpha)^frac1alpha = x^frac1alpha$?







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    No, yes$$.
    – Lord Shark the Unknown
    Jul 18 at 16:32










  • The answer to the title question is "no"
    – Wojowu
    Jul 18 at 16:33










  • Short answer : No.
    – Alex Silva
    Jul 18 at 16:41














up vote
0
down vote

favorite












I want to ask if it is possible to transform equation $a^alpha + b^alpha = x$ into $a + b = x^frac1alpha$ by elevating each parameter by $frac1alpha$? Or if we elevate each parameter by $frac1alpha$ it must be $(a^alpha + b^alpha)^frac1alpha = x^frac1alpha$?







share|cite|improve this question















  • 1




    No, yes$$.
    – Lord Shark the Unknown
    Jul 18 at 16:32










  • The answer to the title question is "no"
    – Wojowu
    Jul 18 at 16:33










  • Short answer : No.
    – Alex Silva
    Jul 18 at 16:41












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I want to ask if it is possible to transform equation $a^alpha + b^alpha = x$ into $a + b = x^frac1alpha$ by elevating each parameter by $frac1alpha$? Or if we elevate each parameter by $frac1alpha$ it must be $(a^alpha + b^alpha)^frac1alpha = x^frac1alpha$?







share|cite|improve this question











I want to ask if it is possible to transform equation $a^alpha + b^alpha = x$ into $a + b = x^frac1alpha$ by elevating each parameter by $frac1alpha$? Or if we elevate each parameter by $frac1alpha$ it must be $(a^alpha + b^alpha)^frac1alpha = x^frac1alpha$?









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share|cite|improve this question




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asked Jul 18 at 16:32









JoeDi

102




102







  • 1




    No, yes$$.
    – Lord Shark the Unknown
    Jul 18 at 16:32










  • The answer to the title question is "no"
    – Wojowu
    Jul 18 at 16:33










  • Short answer : No.
    – Alex Silva
    Jul 18 at 16:41












  • 1




    No, yes$$.
    – Lord Shark the Unknown
    Jul 18 at 16:32










  • The answer to the title question is "no"
    – Wojowu
    Jul 18 at 16:33










  • Short answer : No.
    – Alex Silva
    Jul 18 at 16:41







1




1




No, yes$$.
– Lord Shark the Unknown
Jul 18 at 16:32




No, yes$$.
– Lord Shark the Unknown
Jul 18 at 16:32












The answer to the title question is "no"
– Wojowu
Jul 18 at 16:33




The answer to the title question is "no"
– Wojowu
Jul 18 at 16:33












Short answer : No.
– Alex Silva
Jul 18 at 16:41




Short answer : No.
– Alex Silva
Jul 18 at 16:41










1 Answer
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No it isn't in general since



$$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$



For general $a$ and $b$ it is trivially true only for $alpha=1$.






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    1 Answer
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    active

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    1 Answer
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    active

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    up vote
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    down vote



    accepted










    No it isn't in general since



    $$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$



    For general $a$ and $b$ it is trivially true only for $alpha=1$.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      No it isn't in general since



      $$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$



      For general $a$ and $b$ it is trivially true only for $alpha=1$.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        No it isn't in general since



        $$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$



        For general $a$ and $b$ it is trivially true only for $alpha=1$.






        share|cite|improve this answer













        No it isn't in general since



        $$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$



        For general $a$ and $b$ it is trivially true only for $alpha=1$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 18 at 16:35









        gimusi

        65.4k73584




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