https://file.aiquickdraw.com/mathbot/file/1726123977457o2ytj9in.jpg

To solve this problem, we need to find all pairs of integers \((x, y)\) such that:

\[
\log_x(2 - y) = \log_{y+1}(3 - x).
\]

### Step-by-Step Solution

1. **Express the logarithmic equality in exponential form:**

   Given that \(\log_a(b) = c\) is equivalent to \(a^c = b\), we can rewrite both sides:

   \[
   x^{\log_x(2 - y)} = 2 - y \implies x^A = 2 - y, \quad \text{where } A = \log_x(2 - y),
   \]

   \[
   (y+1)^{\log_{y+1}(3 - x)} = 3 - x \implies (y+1)^B = 3 - x, \quad \text{where } B = \log_{y+1}(3 - x).
   \]

   Since \(A = B\), we have:

   \[
   x^{\log_x(2 - y)} = (y+1)^{\log_{y+1}(3 - x)}.
   \]

   Therefore, we get two expressions:

   \[
   x^A = 2 - y, \quad \text{and} \quad (y+1)^A = 3 - x.
   \]

2. **Analyzing integer solutions:**

   Since both \(x\) and \(y\) are integers, we need to find integer pairs that satisfy the equations:

   \[
   x^A = 2 - y, \quad (y+1)^A = 3 - x.
   \]

   We also need \(x > 0\) and \(y+1 > 0\) for the bases of the logarithms to be valid. This implies:

   \[
   x > 0, \quad y > -1.
   \]

3. **Check possible values systematically:**

   Let's start checking small integer values for \(x\) and \(y\):

   - **Case 1: \(x = 2, y = 1\):**

     Plugging into the equations:

     \[
     \log_2(2 - 1) = \log_2(1) = 0,
     \]

     \[
     \log_{1+1}(3 - 2) = \log_2(1) = 0.
     \]

     Both sides are equal, so \((x, y) = (2, 1)\) is a solution.

4. **Check for other possible solutions:**

   No other small integer values for \(x\) and \(y\) satisfy both conditions because if \(x > 2\) or \(y > 1\), then the values inside the logarithms do not remain valid (logarithm of a non-positive number is undefined).

### Conclusion

The only solution is:

\[
(x, y) = (2, 1).
\]

Would you like to see a deeper exploration of why this is the only solution or any other details?

### Related Questions

1. How would the solution change if \(x\) and \(y\) were not restricted to integers?
2. What are the conditions for the existence of solutions to logarithmic equations in general?
3. How can we solve similar logarithmic equations with different bases?
4. What are some methods to handle inequalities involving logarithms?
5. How can you prove that no other integer pairs satisfy this equation?

### Tip

When solving logarithmic equations, always check the domain constraints to ensure that all expressions inside the logarithms are defined and positive.

Determine all pairs of integers (x, y) such that log_x(2 - y) = log_{y+1}(3 - x). You must use algebra.

Explore how to solve the logarithmic equation log_x(2 - y) = log_{y+1}(3 - x) by determining all integer pairs (x, y) that satisfy this equation. This problem covers logarithms, algebra, and exponential equations, making it ideal for 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