Transform exponential expression into log / find solution to: $y = (a - x_1)^gamma + (a - x_2)^gamma$

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I need to find the solution to the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ for $a$, but have problems with exponential, where $gamma $ can be greater than 0 and smaller than 1 or greater than 1: (1) $0 < gamma < 1$ and (2) $gamma > 1$. I think $(a - x_1)^gamma$ has to be to transformed into logarithm, but I can't find the way to solve it. I would appreciate any hint or help.




linear-algebra logarithms exponential-function




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  • But log of sum isn't sum of logs...
    – coffeemath
    Jul 18 at 12:20






  • 1




    what is the unknown here? Is it just $gamma$?
    – Vasya
    Jul 18 at 12:53










  • I need to solve it for $a$.
    – JoeDi
    Jul 18 at 13:11














up vote
0
down vote

favorite












I need to find the solution to the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ for $a$, but have problems with exponential, where $gamma $ can be greater than 0 and smaller than 1 or greater than 1: (1) $0 < gamma < 1$ and (2) $gamma > 1$. I think $(a - x_1)^gamma$ has to be to transformed into logarithm, but I can't find the way to solve it. I would appreciate any hint or help.




linear-algebra logarithms exponential-function




share|cite|improve this question





















  • But log of sum isn't sum of logs...
    – coffeemath
    Jul 18 at 12:20






  • 1




    what is the unknown here? Is it just $gamma$?
    – Vasya
    Jul 18 at 12:53










  • I need to solve it for $a$.
    – JoeDi
    Jul 18 at 13:11












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I need to find the solution to the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ for $a$, but have problems with exponential, where $gamma $ can be greater than 0 and smaller than 1 or greater than 1: (1) $0 < gamma < 1$ and (2) $gamma > 1$. I think $(a - x_1)^gamma$ has to be to transformed into logarithm, but I can't find the way to solve it. I would appreciate any hint or help.




linear-algebra logarithms exponential-function




share|cite|improve this question













I need to find the solution to the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ for $a$, but have problems with exponential, where $gamma $ can be greater than 0 and smaller than 1 or greater than 1: (1) $0 < gamma < 1$ and (2) $gamma > 1$. I think $(a - x_1)^gamma$ has to be to transformed into logarithm, but I can't find the way to solve it. I would appreciate any hint or help.





linear-algebra logarithms exponential-function





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 18 at 13:11
























asked Jul 18 at 12:08









JoeDi

102




102











  • But log of sum isn't sum of logs...
    – coffeemath
    Jul 18 at 12:20






  • 1




    what is the unknown here? Is it just $gamma$?
    – Vasya
    Jul 18 at 12:53










  • I need to solve it for $a$.
    – JoeDi
    Jul 18 at 13:11
















  • But log of sum isn't sum of logs...
    – coffeemath
    Jul 18 at 12:20






  • 1




    what is the unknown here? Is it just $gamma$?
    – Vasya
    Jul 18 at 12:53










  • I need to solve it for $a$.
    – JoeDi
    Jul 18 at 13:11















But log of sum isn't sum of logs...
– coffeemath
Jul 18 at 12:20




But log of sum isn't sum of logs...
– coffeemath
Jul 18 at 12:20




1




1




what is the unknown here? Is it just $gamma$?
– Vasya
Jul 18 at 12:53




what is the unknown here? Is it just $gamma$?
– Vasya
Jul 18 at 12:53












I need to solve it for $a$.
– JoeDi
Jul 18 at 13:11




I need to solve it for $a$.
– JoeDi
Jul 18 at 13:11










1 Answer
1






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If I can assume $a>x_1>x_2$, which is ok if $x_1-x_2<y^1/gamma$, then $a$ is between $x_1$ and $y^1/gamma+x_2$. You can then try the Bisection method to home in on the answer.






share|cite|improve this answer





















  • Sorry for my low level of math and maybe a stupid question, but so we can elevate the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ by $frac1gamma$ and get $y^frac1gamma = (a - x_1) + (a - x_2)$?
    – JoeDi
    Jul 18 at 14:22










  • No. But when $a=x_2+y^1/gamma$, the right-hand side is more than $(a-x_2)^gamma=(y^1/gamma)^gamma=y$
    – Empy2
    Jul 18 at 21:16










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










If I can assume $a>x_1>x_2$, which is ok if $x_1-x_2<y^1/gamma$, then $a$ is between $x_1$ and $y^1/gamma+x_2$. You can then try the Bisection method to home in on the answer.






share|cite|improve this answer





















  • Sorry for my low level of math and maybe a stupid question, but so we can elevate the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ by $frac1gamma$ and get $y^frac1gamma = (a - x_1) + (a - x_2)$?
    – JoeDi
    Jul 18 at 14:22










  • No. But when $a=x_2+y^1/gamma$, the right-hand side is more than $(a-x_2)^gamma=(y^1/gamma)^gamma=y$
    – Empy2
    Jul 18 at 21:16














up vote
0
down vote



accepted










If I can assume $a>x_1>x_2$, which is ok if $x_1-x_2<y^1/gamma$, then $a$ is between $x_1$ and $y^1/gamma+x_2$. You can then try the Bisection method to home in on the answer.






share|cite|improve this answer





















  • Sorry for my low level of math and maybe a stupid question, but so we can elevate the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ by $frac1gamma$ and get $y^frac1gamma = (a - x_1) + (a - x_2)$?
    – JoeDi
    Jul 18 at 14:22










  • No. But when $a=x_2+y^1/gamma$, the right-hand side is more than $(a-x_2)^gamma=(y^1/gamma)^gamma=y$
    – Empy2
    Jul 18 at 21:16












up vote
0
down vote



accepted







up vote
0
down vote



accepted






If I can assume $a>x_1>x_2$, which is ok if $x_1-x_2<y^1/gamma$, then $a$ is between $x_1$ and $y^1/gamma+x_2$. You can then try the Bisection method to home in on the answer.






share|cite|improve this answer













If I can assume $a>x_1>x_2$, which is ok if $x_1-x_2<y^1/gamma$, then $a$ is between $x_1$ and $y^1/gamma+x_2$. You can then try the Bisection method to home in on the answer.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 18 at 13:34









Empy2

31.9k12059




31.9k12059











  • Sorry for my low level of math and maybe a stupid question, but so we can elevate the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ by $frac1gamma$ and get $y^frac1gamma = (a - x_1) + (a - x_2)$?
    – JoeDi
    Jul 18 at 14:22










  • No. But when $a=x_2+y^1/gamma$, the right-hand side is more than $(a-x_2)^gamma=(y^1/gamma)^gamma=y$
    – Empy2
    Jul 18 at 21:16
















  • Sorry for my low level of math and maybe a stupid question, but so we can elevate the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ by $frac1gamma$ and get $y^frac1gamma = (a - x_1) + (a - x_2)$?
    – JoeDi
    Jul 18 at 14:22










  • No. But when $a=x_2+y^1/gamma$, the right-hand side is more than $(a-x_2)^gamma=(y^1/gamma)^gamma=y$
    – Empy2
    Jul 18 at 21:16















Sorry for my low level of math and maybe a stupid question, but so we can elevate the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ by $frac1gamma$ and get $y^frac1gamma = (a - x_1) + (a - x_2)$?
– JoeDi
Jul 18 at 14:22




Sorry for my low level of math and maybe a stupid question, but so we can elevate the expression $y = (a - x_1)^gamma + (a - x_2)^gamma$ by $frac1gamma$ and get $y^frac1gamma = (a - x_1) + (a - x_2)$?
– JoeDi
Jul 18 at 14:22












No. But when $a=x_2+y^1/gamma$, the right-hand side is more than $(a-x_2)^gamma=(y^1/gamma)^gamma=y$
– Empy2
Jul 18 at 21:16




No. But when $a=x_2+y^1/gamma$, the right-hand side is more than $(a-x_2)^gamma=(y^1/gamma)^gamma=y$
– Empy2
Jul 18 at 21:16












 

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