# if (x^2)+(y^2)=34xy , show that log[(x+y)/6]=1/2(log x + log y)

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You should remember that `x^2 + y^2 = (x + y)^2 - 2xy` , hence, you need to substitute `(x + y)^2 - 2xy` for `x^2 + y^2` in the given condition such that:

`(x + y)^2 - 2xy = 34xy =gt (x + y)^2 = 34xy + 2xy`

`(x + y)^2 = 36xy`

You need to take logarithms both sides such that:

`log (x + y)^2 = log (36xy)`

`2log (x + y) = log 36 + log xy`

`2log (x + y) - log36 = log xy`

You may substitute `6^2` for 36 such that:

`2log (x + y) - log 6^2 = log xy`

Factoring out 2 yields:

`2(log (x + y) - log 6) = log xy`

You need to use the logarithmic identities, hence, you need to convert the difference of logarithm from the left, to the logarithm of quotient, such that:

`2(log ((x + y)/6)) = log xy`

You need to convert the logarithm of product into the sum of logarithms such that:

`2(log ((x + y)/6)) = log x + log y`

You need to divide by 2 such that:

`(log ((x + y)/6)) = (1/2)(log x + log y)`

**Hence, using the logarithms in the given relation `x^2 + y^2 = 34xy` yields `(log ((x + y)/6)) = (1/2)(log x + log y)` .**

**Sources:**