Answer:

To prove that \(\log(x) + \log(y) + \log(3) + 2\log(2) = 2\log(x + y)\), we'll start with the given equation \(x^2 + y^2 = 10xy\).

First, divide both sides by \(xy\):

\(\frac{x^2}{xy} + \frac{y^2}{xy} = \frac{10xy}{xy}\).

This simplifies to:

\(x + y = 10\).

Now, let's find the value of \(\log(x + y)\):

\(\log(x + y) = \log(10)\).

Using logarithm properties, we can express \(\log(10)\) as \(\log(2 \cdot 5)\):

\(\log(10) = \log(2) + \log(5)\).

Now, we can substitute the values of \(\log(2)\) and \(\log(5)\) using the properties of logarithms:

\(\log(x + y) = \log(2) + \log(5) = \log(2) + \log(3) + \log(2)\).

Finally, simplify the right side:

\(\log(2) + \log(3) + \log(2) = 2\log(2) + \log(3) = 2\log(2) + \log(3)\).

So, \(\log(x) + \log(y) + \log(3) + 2\log(2) = 2\log(x + y)\) is indeed proven.