If `2^(x+y)=6^(y)` and `3^(x-1)=2^(y+1)`, then the value of `(log3-log2)//(x-y)` is

Options

A. `1`
B. `log_(2)3-log_(3)2`
C. `log(3//2)`
D. `log3-log2`

Answer

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

1. \( 2^{(x+y)} = 6^y \)
2. \( 3^{(x-1)} = 2^{(y+1)} \)

We need to find the value of \( \frac{\log 3 - \log 2}{x - y} \).

### Step 1: Take logarithms of both equations

For the first equation:

\[ \log(2^{(x+y)}) = \log(6^y) \]

Using the property of logarithms \( \log(a^b) = b \log(a) \):

\[ (x+y) \log 2 = y \log 6 \]

Now, we can express \( \log 6 \) as \( \log(2 \cdot 3) = \log 2 + \log 3 \):

\[ (x+y) \log 2 = y (\log 2 + \log 3) \]

Expanding this gives:

\[ x \log 2 + y \log 2 = y \log 2 + y \log 3 \]

Subtracting \( y \log 2 \) from both sides:

\[ x \log 2 = y \log 3 \]

Thus, we have:

\[ \frac{x}{y} = \frac{\log 3}{\log 2} \quad \text{(1)} \]

### Step 2: Take logarithms of the second equation

For the second equation:

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

Using the property of logarithms:

\[ (x-1) \log 3 = (y+1) \log 2 \]

Expanding this gives:

\[ x \log 3 - \log 3 = y \log 2 + \log 2 \]

Rearranging gives:

\[ x \log 3 - y \log 2 = \log 3 + \log 2 \]

Now we can express this as:

\[ x \log 3 = y \log 2 + \log 3 + \log 2 \]

### Step 3: Substitute \( x \) from equation (1)

From equation (1), we can express \( x \) in terms of \( y \):

\[ x = y \cdot \frac{\log 3}{\log 2} \]

Substituting this into the equation:

\[ \left(y \cdot \frac{\log 3}{\log 2}\right) \log 3 = y \log 2 + \log 3 + \log 2 \]

### Step 4: Simplify the equation

Expanding gives:

\[ y \cdot \frac{\log 3^2}{\log 2} = y \log 2 + \log 3 + \log 2 \]

Multiplying through by \( \log 2 \) to eliminate the fraction:

\[ y \log 3^2 = y \log^2 2 + \log 2 \log 3 + \log^2 2 \]

### Step 5: Solve for \( y \)

Rearranging gives:

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

Thus:

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

### Step 6: Find \( x - y \)

Using the expression for \( x \):

\[ x - y = y \cdot \frac{\log 3}{\log 2} - y \]

Factoring out \( y \):

\[ x - y = y \left( \frac{\log 3}{\log 2} - 1 \right) = y \cdot \frac{\log 3 - \log 2}{\log 2} \]

### Step 7: Find \( \frac{\log 3 - \log 2}{x - y} \)

Substituting into our expression:

\[ \frac{\log 3 - \log 2}{x - y} = \frac{\log 3 - \log 2}{y \cdot \frac{\log 3 - \log 2}{\log 2}} = \frac{\log 2}{y} \]

### Final Step: Substitute \( y \)

We know \( y \) from earlier, hence:

\[ \frac{\log 2}{y} = \frac{\log 2}{\frac{\log 2 \log 3 + \log^2 2}{\log 3^2 - \log^2 2}} = \frac{(\log 3^2 - \log^2 2) \log 2}{\log 2 \log 3 + \log^2 2} \]

This simplifies to:

\[ \frac{\log 3^2 - \log^2 2}{\log 3 + \log 2} \]

Thus, the final answer is:

\[ \frac{\log 3 - \log 2}{x - y} = 1 \]