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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
asked Jul 18 at 12:08
JoeDi
102
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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
asked Jul 18 at 12:08
JoeDi
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
add a comment |Â
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
asked Jul 18 at 12:08
JoeDi
102
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
asked Jul 18 at 12:08
JoeDi
102
edited Jul 18 at 13:11
asked Jul 18 at 12:08
JoeDi
102
asked Jul 18 at 12:08
JoeDi
102
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
add a comment |Â
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
add a comment |Â
1 Answer
1
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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.
answered Jul 18 at 13:34
Empy2
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
add a comment |Â
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.
answered Jul 18 at 13:34
Empy2
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
add a comment |Â
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.
answered Jul 18 at 13:34
Empy2
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
add a comment |Â
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.
answered Jul 18 at 13:34
Empy2
31.9k12059
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.
answered Jul 18 at 13:34
Empy2
31.9k12059
answered Jul 18 at 13:34
Empy2
31.9k12059
answered Jul 18 at 13:34
Empy2
31.9k12059
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
add a comment |Â
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
add a comment |Â
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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