What are x and y if 4^(x/y)*4^(y/x)=32 and log 3 (x-y)=1-log 3 (x+y) ?

We'll work on the 1st equation. We notice that the bases
of the multiplied factors are matching, so we'll add the
exponents:

4^(x/y)*4^(y/x) = 4^(x/y + y/x) = 4^[(x^2 +
y^2)/xy]

But 4 = 2^2 and we'll raise to the exponent [(x^2
+ y^2)/xy], both sides:

4^[(x^2 + y^2)/xy] = 2^2*[(x^2 +
y^2)/xy]

32 = 2^5

The 1st
equation will become:

2^2*[(x^2 + y^2)/xy] =
2^5

Since the bases are matching, we'll apply one to one
property:

2*[(x^2 + y^2)/xy] =
5

We'll note x/y = z => y/x =
1/z

2(z + 1/z) = 5

2z^2 - 5z +
2 = 0

We'll determine the
roots:

z1 = [5+sqrt(25 -
16)]/4

z1 = (5+3)/4

z1 =
2

z2 = 1/2

We'll put x/y = z1
<=> x/y = 2 => x = 2y

x/y = 1/2
=> y = 2x

We'll re-write the 2nd equation, using the
product rule of logarithms:

log3 (x-y) + log3 (x+y) =
1

log3 (x^2 - y^2) = 1

We'll
take antilogarithms:

x^2 - y^2 =
3

We'll put x = 2y:

4y^2 - y^2
= 3

3y^2 = 3

y^2 =
1

y1 = 1 => x = 2

y2 =
-1 => x = -2

We'll put x =
y/2:

y^2/4 - y^2 = 3

y^2 -
4y^2 = 12

-3y^2 = 12

y^2 =
-4

y1 = 2i => x1 = 2i/2 =
i

y2 = -2i => x =
-i

Since x and y values have to be real for
the logarithms to exist, we'll accept only the real solutions of the equation. 
Furthermore, the logarithms must be positive and the only solution that satifies all
these requirements is (2, 1)