If `log_(2)(x + y) = log_(3)(x - y) = (log 25)/(log 0.2)`, find the values of x and y.

Answer

To solve the problem, we start with the given equations:

1. \( \log_{2}(x + y) = \log_{3}(x - y) = \frac{\log 25}{\log 0.2} \)

### Step 1: Simplify \( \frac{\log 25}{\log 0.2} \)

Using the change of base formula, we can simplify \( \frac{\log 25}{\log 0.2} \):

\[ \log 0.2 = \log \left(\frac{2}{10}\right) = \log 2 - \log 10 = \log 2 - 1 \]

Thus,

\[ \frac{\log 25}{\log 0.2} = \frac{\log 25}{\log 2 - 1} \]

Since \( \log 25 = \log (5^2) = 2 \log 5 \), we can rewrite it as:

\[ \frac{2 \log 5}{\log 2 - 1} \]

### Step 2: Set the equations

Now, we have:

\[ \log_{2}(x + y) = \log_{3}(x - y) = \frac{2 \log 5}{\log 2 - 1} \]

Let’s denote this common value as \( k \):

\[ k = \frac{2 \log 5}{\log 2 - 1} \]

### Step 3: Convert logarithmic equations to exponential form

From \( \log_{2}(x + y) = k \):

\[ x + y = 2^k \]

From \( \log_{3}(x - y) = k \):

\[ x - y = 3^k \]

### Step 4: Solve the system of equations

We now have two equations:

1. \( x + y = 2^k \) (Equation 1)
2. \( x - y = 3^k \) (Equation 2)

Adding these two equations:

\[ (x + y) + (x - y) = 2^k + 3^k \]

This simplifies to:

\[ 2x = 2^k + 3^k \]

Thus,

\[ x = \frac{2^k + 3^k}{2} \]

Subtracting Equation 2 from Equation 1:

\[ (x + y) - (x - y) = 2^k - 3^k \]

This simplifies to:

\[ 2y = 2^k - 3^k \]

Thus,

\[ y = \frac{2^k - 3^k}{2} \]

### Step 5: Substitute \( k \)

Now we need to substitute \( k \) back into the equations for \( x \) and \( y \):

1. \( x = \frac{2^{\frac{2 \log 5}{\log 2 - 1}} + 3^{\frac{2 \log 5}{\log 2 - 1}}}{2} \)
2. \( y = \frac{2^{\frac{2 \log 5}{\log 2 - 1}} - 3^{\frac{2 \log 5}{\log 2 - 1}}}{2} \)

### Final Values

Now, we can compute \( x \) and \( y \) using the values of \( k \) we derived.

### Summary of Results

After calculating, we find:

- \( x = \frac{13}{72} \)
- \( y = \frac{5}{72} \)