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

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### 2 Answers

We have the equations:

4^(x/y)*4^(y/x) = 32 ...(1)

log 3 (x-y) = 1 - log 3 (x+y) ...(2)

From (1) we get

4^(x/y)*4^(y/x) = 32

=> 4^(x/y + y/x) = 2^5

=> 2^2(x/y + y/x) = 2^5

equate the exponents as the base is the same

2(x/y + y/x) = 5

=> x/y + y/x = 2.5

We can get this using x/y = 1/2 and y/x = 2

The solution sets we get are (1,2) , (2,1) (-1,-2) and (-2, -1)

But log(3)(x - y) = 1 - log(3)(x + y), so x- y and x + y should be positive. This is possible only with the solution (2, 1) as with all the other sets either x - y or x + y is negative.

**The solution of the equation is only (2,1)**

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)**