# math2^(x+y) = 32 x*y=6 Determine x and y

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You may solve the top equation using the logarithmation operation, such that:

`log_2 2^(x+y) = log_2 32 `

Using logarithimc identities, yields:

`(x + y)log_2 2 = 5log_2 2 => x + y = 5 = p`

Using the bottom equation, `xy = 6 = q` , and Lagrange resolvents yields:

`x^2 - px + q = 0 => x^2 - 5x + 6 = 0`

You may solve quadratic equation completing the square, such that:

`x^2 - 5x + (5/2)^2 = (5/2)^2 - 6`

`(x - 5/2)^2 = (25 - 24)/4`

`x - 5/2 = +-sqrt(1/4) => x_1 = 5/2 + 1/2 => x_1 = 3`

`x_2 = 5/2 - 1/2 => x_2 = 2`

**Hence, evaluating the solutions to simultaneous equations yields `x = 3, y = 2` or **`x = 2, y = 3.`

It is given that 2^(x+y) = 32 and x*y=6 . This has to be used to solve for x and y.

From x*y = 6 we get x = 6/y

In the equation 2^(x + y) = 32 make the base equal for both the sides. 32 = 2^5

2^(x + y) = 2^5

Substitute x = 6/y

2^(6/y + y) = 2^5

Equating the exponents 6/y + y = 5

y^2 - 5y + 6 = 0

y^2 - 3y - 2y + 6 = 0

y(y - 3) - 2(y - 3) = 0

(y - 2)(y - 3) = 0

y = 2 and y = 3

The solution of the given system of equations is x = 2, y=3 and x = 3, y = 2

We notice that 32 is a power of 2 and we could create matching bases in the 1st equation.

2^(x+y) =2^5

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

x + y = 5 (1)

x*y=6 (2)

We'll note x + y = S and x*y = P.

S = 5 and P = 6

We'll create the quadratic equation with the sum and the product;

x^2 - Sx + P = 0

x^2 - 5x + 6 = 0

We'll apply the quadratic formula:

x1 = [5+sqrt(25 - 24)]/2

x1 = (5 + 1)/2

x1 = 3

x2 = 2

So, the system of equations will have the following solutions {2;3} and {3;2}.