Let's analyze and solve the system of equations presented in the image:

1. The given equations are: $\log_2(y + 1) + \log_2\left(\frac{1}{2}x\right) = 3$ $\log_2(y - 1) + \log_2(x) = 2$

Step 1: Simplify the first equation.

We use the logarithmic identity: $\log_b(A) + \log_b(B) = \log_b(A \cdot B)$

Thus, for the first equation: $\log_2\left((y + 1) \cdot \frac{1}{2}x\right) = 3$ Simplifying further: $\log_2\left(\frac{x(y + 1)}{2}\right) = 3$

Now, by the definition of logarithms: $\frac{x(y + 1)}{2} = 2^3$ $\frac{x(y + 1)}{2} = 8$ Multiplying both sides by 2: x(y + 1) = 16 \tag{1}

Step 2: Simplify the second equation.

Using the same logarithmic identity for the second equation: $\log_2((y - 1) \cdot x) = 2$ By the definition of logarithms: $(y - 1)x = 2^2$ (y - 1)x = 4 \tag{2}

Step 3: Solve the system of equations.

We now have the system:

1. $x(y + 1) = 16$
2. $(y - 1)x = 4$

From equation (2): $x = \frac{4}{y - 1}$ Substitute this expression for $x$ into equation (1): $\frac{4}{y - 1}(y + 1) = 16$ Multiply both sides by $y - 1$: $4(y + 1) = 16(y - 1)$ Expand both sides: $4y + 4 = 16y - 16$ Simplify: $4 + 16 = 16y - 4y$ $20 = 12y$ $y = \frac{20}{12} = \frac{5}{3}$

Step 4: Solve for $x$.

Substitute $y = \frac{5}{3}$ into equation (2): $x = \frac{4}{y - 1} = \frac{4}{\frac{5}{3} - 1} = \frac{4}{\frac{5}{3} - \frac{3}{3}} = \frac{4}{\frac{2}{3}} = 4 \times \frac{3}{2} = 6$

Final Answer:

• $x = 6$
• $y = \frac{5}{3}$

Would you like a deeper explanation of any step or further clarifications?

Here are some related questions you might want to explore:

1. How do logarithmic identities simplify solving equations?
2. Can logarithmic functions have negative arguments? Why or why not?
3. What would change if the base of the logarithms were different?
4. How would you graph the system of equations involving logarithms?
5. What are the properties of solutions to logarithmic equations?

Tip: Always check the domain of logarithmic equations to ensure the arguments are positive!