# Gven x^4+y^4=641 and logx+logy=1, what is x+y?

*print*Print*list*Cite

### 1 Answer

You need to use the information provided by the problem to evaluate the summation` x + y` , hence, you need first to convert the summation of logarithms into the logarithm of product, such that:

`log x + log y = log (x*y) => log (x*y) = 1 => x*y = 10^1 => x*y = 10`

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

`(x^2 + y^2)^2 = x^4 + 2x^2*y^2 + y^4`

Keeping `x^4 + y^4` to one side, yields:

`x^4 + y^4 = (x^2 + y^2)^2 - 2x^2*y^2`

Since `x*y = 10` , you may replace `(x*y)^2 = 10^2` = `100`

, such that:

`x^4 + y^4 = (x^2 + y^2)^2 -200`

Replacing `(x + y)^2 - 2xy` for `x^2 + y^2` , yields:

`x^4 + y^4 = ((x + y)^2 - 2xy)^2 - 200`

`x^4 + y^4 = ((x + y)^2 - 2*10)^2 - 200`

Replacing `641` for `x^4 + y^4` yields:

`641 = ((x + y)^2 - 20)^2 - 200 => ((x + y)^2 - 20)^2 = 841 `

Taking square roots both sides, yields:

`sqrt (((x + y)^2 - 20)^2) = sqrt 841`

`|(x + y)^2 - 20| = 29 => (x + y)^2 - 20 = +-29`

Considering `(x + y)^2 - 20 = 29` yields:

`(x + y)^2 - 20 = 29 => (x + y)^2 = 49 => x + y = +-7`

Since x and y are the arguments of logarithms, hence, they need to be both positive, thus, its summation needs to be positive. You need to keep only the positive value `x + y = 7.`

**Hence, evaluating the summation `x + y,` under the given conditions, yields **`x + y = 7.`