# What is solution in equations? 9^(x^1/3)+4^(y^1/4)=97 ; 3^(x^1/3)2^(y^1/4)=36

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### 1 Answer

You need to solve the given system of equations, hence, it is recommandable for you to perform the following substitution, such that:

`{(3^(root(3)x) = u),(2^(root(4)y) = v):} => {(9^(root(3)x) = u^2),(4^(root(4)y) = v^2):}`

Replacing the variables x and y with u and v, yields:

`{(u^2 + v^2 = 97),(u*v = 36):}`

You need to use the following special product, such that:

`(u + v)^2 - 2uv = u^2 + v^2 `

Replacing `(u + v)^2 - 2uv` for `u^2 + v^2` yields:

`{((u + v)^2 - 2uv = 97),(u*v = 36):}`

Replacing 36 for u*v yields:

`{((u + v)^2 - 2*36 = 97),(u*v = 36):}`

`{((u + v)^2 - 72 = 97),(u*v = 36):}`

`{((u + v)^2 = 97 + 72),(u*v = 36):}`

`{((u + v)^2 = 169),(u*v = 36):}`

`{(u + v = +-sqrt169),(u*v = 36):}`

`{(u + v = +-13),(u*v = 36):}`

You should use Lagrange's resolvents, such that:

`z^2 - (u+v)*z + u*v = 0`

`z^2 - 13z + 36 = 0`

Using quadratic formula yields:

`z_(1,2) = (13+-sqrt(169 - 144))/2`

`z_(1,2) = (13+-5)/2 => z_1 = 9 ; z_2 = 4 => {(u = 9),(v = 4):} or {(u = 4),(v = 9):} `

`z_(3,4) = (-13+-5)/2 => z_3 = -4 ; z_2 = -9 => {(u = -9),(v = -4):} or {(u = -4),(v = -9):}`

Replacing back `3^(root(3)x)` for u and `2^(root(4)y)` for v yields:

`3^(root(3)x) = 9 => root(3)x = 2 => x = 2^3 => x = 8`

`3^(root(3)x) = 4 => root(3)x*ln 3 = ln 4 => x = ((ln 4)/(ln 3))^3`

`3^(root(3)x) = -9` invalid

`3^(root(3)x) = -4` invalid

`2^(root(4)y) = 4 => root(4)y = 2 => y = 2^4 => y = 16`

`2^(root(4)y) = 9 => root(4)y*ln 2 = ln 9 => y = ((ln 9)/(ln 2))^4`

`2^(root(4)y) = -4` invalid

`2^(root(4)y) = -9` invalid

**Hence, evaluating the solutions to the system of equations yields `x = 8, y = 16` and **`x = ((ln 4)/(ln 3))^3, y = ((ln 9)/(ln 2))^4.`

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