We are given the equation:

$\ln\left(\frac{2}{-y + 3}\right) = \ln y$

To solve for $y$, we can take the following steps:

Step 1: Use the property of logarithms

Since the natural logarithm ($\ln$) function is one-to-one, we can set the arguments inside the logarithms equal to each other:

$\frac{2}{-y + 3} = y$

Step 2: Multiply both sides by $-y + 3$ to eliminate the fraction:

$2 = y(-y + 3)$

Step 3: Expand the right-hand side:

$2 = -y^2 + 3y$

Step 4: Rearrange the equation into standard quadratic form:

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

Step 5: Factor the quadratic equation:

$(y - 1)(y - 2) = 0$

Step 6: Solve for $y$:

$y = 1 \quad \text{or} \quad y = 2$

Step 7: Check for extraneous solutions

We need to check if both values of $y$ satisfy the original equation. Since the natural logarithm is only defined for positive values, we discard any negative or zero results from the argument of $\ln y$.

• For $y = 1$: $\ln\left(\frac{2}{-1 + 3}\right) = \ln 1$ which gives $\ln 1 = \ln 1$, a true statement.

• For $y = 2$: $\ln\left(\frac{2}{-2 + 3}\right) = \ln 2$ which gives $\ln 2 = \ln 2$, a true statement.

Conclusion:

Both $y = 1$ and $y = 2$ are valid solutions.

Would you like a more detailed explanation, or do you have any questions?

Here are some related questions to explore:

1. What happens if we have $\ln\left(\frac{a}{b}\right) = \ln x$ in general?
2. How do you solve equations with multiple logarithms on each side?
3. What are the conditions for a logarithmic function to be defined?
4. How would this equation change if the logarithm base were different?
5. What if we introduce a constant outside the logarithms, like $c \cdot \ln\left(\frac{2}{-y + 3}\right) = \ln y$?

Tip: Always remember to check the domain of the logarithmic function when solving logarithmic equations, as logarithms are only defined for positive numbers.