We are given the system of equations:

1. $x + y = 8$
2. $\log_2(xy) = 2$

We are tasked with finding $y$.

Step 1: Express $xy$ from the logarithmic equation

From the second equation, $\log_2(xy) = 2$, we can rewrite it in exponential form: $xy = 2^2 = 4$ So, we now have the system:

1. $x + y = 8$
2. $xy = 4$

Step 2: Solve for $y$

We can solve for $y$ in terms of $x$ from the first equation: $y = 8 - x$

Step 3: Substitute into the second equation

Substitute $y = 8 - x$ into the second equation $xy = 4$: $x(8 - x) = 4$ Simplifying: $8x - x^2 = 4$ Rearranging into standard quadratic form: $x^2 - 8x + 4 = 0$

Step 4: Solve the quadratic equation

We can solve the quadratic equation $x^2 - 8x + 4 = 0$ using the quadratic formula: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(4)}}{2(1)}$ $x = \frac{8 \pm \sqrt{64 - 16}}{2}$ $x = \frac{8 \pm \sqrt{48}}{2}$ $x = \frac{8 \pm 4\sqrt{3}}{2}$ $x = 4 \pm 2\sqrt{3}$

Step 5: Find the corresponding $y$

Now that we have two possible values for $x$:

1. $x = 4 + 2\sqrt{3}$
2. $x = 4 - 2\sqrt{3}$

We can find the corresponding values for $y$ using $y = 8 - x$:

• If $x = 4 + 2\sqrt{3}$, then: $y = 8 - (4 + 2\sqrt{3}) = 4 - 2\sqrt{3}$
• If $x = 4 - 2\sqrt{3}$, then: $y = 8 - (4 - 2\sqrt{3}) = 4 + 2\sqrt{3}$

Thus, the two possible values for $y$ are: $y = 4 - 2\sqrt{3} \quad \text{or} \quad y = 4 + 2\sqrt{3}$

Final Answer:

The possible values for $y$ are $\boxed{4 - 2\sqrt{3}}$ and $\boxed{4 + 2\sqrt{3}}$.

Would you like further details or clarifications? Here are a few related questions:

1. How do we verify the solution using the logarithmic equation?
2. How can we solve a quadratic equation using factoring?
3. What is the significance of logarithmic equations in solving systems like this?
4. Can we have negative values for $x$ and $y$ in this scenario?
5. How does the quadratic formula work in this type of problem?

Tip: Always check for extraneous solutions when solving logarithmic and exponential equations, especially if the variable is inside a logarithm.