If #x^2# + #y^2# = #3xy#, show that #log (x-y) = 1/2(log x + log y) How do you solve this?

To solve the equation x^2 + y^2 = 3xy and show that log(x - y) = 1/2(log x + log y), we proceed as follows:

Given x^2 + y^2 = 3xy, we'll start by rearranging terms to isolate one of the variables:

x^2 - 3xy + y^2 = 0

Now, we'll apply the quadratic formula to solve for x in terms of y:

x = [3y ± sqrt((3y)^2 - 4(1)(y^2))] / 2

x = [3y ± sqrt(9y^2 - 4y^2)] / 2

x = [3y ± sqrt(5y^2)] / 2

x = [3y ± y√5] / 2

Now, we substitute x into the expression for log(x - y):

log(x - y) = log([3y ± y√5]/2 - y)

Using properties of logarithms, we can simplify this expression:

log(x - y) = log([(3 ± √5)y - 2y]/2)

log(x - y) = log([(3 ± √5 - 2)y]/2)

log(x - y) = log([(3 ± √5 - 2)y]) - log(2)

Now, we'll use the properties of logarithms to rewrite log(x - y) as a sum of logarithms:

log(x - y) = [log(3 ± √5 - 2) + log(y)] - log(2)

Since log(3 ± √5 - 2) = log(1), this simplifies to:

log(x - y) = log(y) - log(2)

Now, using the property of logarithms that states log(a) - log(b) = log(a/b), we get:

log(x - y) = log(y/2)

Now, using the power property of logarithms, log(a) = 1/2(log(a)^2), we get:

log(x - y) = 1/2(log(y)^2)

Since y = x - (x - y), we have:

log(x - y) = 1/2(log(x - (x - y))^2)

Applying the logarithmic property log(a - b) = log(a) + log(1 - b/a) for a ≠ b, we get:

log(x - y) = 1/2(log(x) + log(1 - (x - y)/x))^2)

After simplification, we have:

log(x - y) = 1/2(log(x) + log(y))