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.

Learn how to solve the logarithmic equation ln(2/(-y + 3)) = ln(y) with a detailed, step-by-step solution. This problem covers logarithmic properties, algebraic manipulation, and solving quadratic equations. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation