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. none of these

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 \( ( \log 3 - \log 2 )( x - y ) \).

### Step 1: Rewrite the first equation using logarithms

Taking the logarithm (base 2) of both sides of the first equation:

\[ \log_2(2^{x+y}) = \log_2(6^y) \]

Using the property of logarithms \( \log_b(a^n) = n \log_b(a) \):

\[ x + y = y \log_2(6) \]

### Step 2: Rewrite \( \log_2(6) \)

We can express \( 6 \) as \( 2 \cdot 3 \):

\[ \log_2(6) = \log_2(2 \cdot 3) = \log_2(2) + \log_2(3) = 1 + \log_2(3) \]

Substituting this back into our equation gives:

\[ x + y = y(1 + \log_2(3)) \]

### Step 3: Rearranging the first equation

Rearranging the equation:

\[ x + y = y + y \log_2(3) \]

This simplifies to:

\[ x = y \log_2(3) \]

### Step 4: Rewrite the second equation using logarithms

Now, take the logarithm (base 2) of both sides of the second equation:

\[ \log_2(3^{x-1}) = \log_2(2^{y+1}) \]

Using the property of logarithms:

\[ (x-1) \log_2(3) = (y+1) \log_2(2) \]

Since \( \log_2(2) = 1 \):

\[ (x-1) \log_2(3) = y + 1 \]

### Step 5: Rearranging the second equation

Rearranging gives:

\[ x \log_2(3) - \log_2(3) = y + 1 \]

This simplifies to:

\[ x \log_2(3) - y = \log_2(3) + 1 \]

### Step 6: Substitute \( x \) from the first equation

Now substitute \( x = y \log_2(3) \) into the equation:

\[ (y \log_2(3)) \log_2(3) - y = \log_2(3) + 1 \]

This expands to:

\[ y (\log_2(3))^2 - y = \log_2(3) + 1 \]

### Step 7: Factor out \( y \)

Factoring out \( y \):

\[ y \left( (\log_2(3))^2 - 1 \right) = \log_2(3) + 1 \]

### Step 8: Solve for \( y \)

Thus:

\[ y = \frac{\log_2(3) + 1}{(\log_2(3))^2 - 1} \]

### Step 9: Substitute \( y \) back to find \( x \)

Now substitute \( y \) back into \( x = y \log_2(3) \):

\[ x = \frac{(\log_2(3) + 1) \log_2(3)}{(\log_2(3))^2 - 1} \]

### Step 10: Find \( x - y \)

Now, we need \( x - y \):

\[ x - y = \frac{(\log_2(3) + 1) \log_2(3)}{(\log_2(3))^2 - 1} - \frac{\log_2(3) + 1}{(\log_2(3))^2 - 1} \]

This simplifies to:

\[ x - y = \frac{(\log_2(3) + 1)(\log_2(3) - 1)}{(\log_2(3))^2 - 1} \]

### Step 11: Calculate \( ( \log 3 - \log 2 )( x - y ) \)

Now we can compute:

\[ ( \log 3 - \log 2 )( x - y ) \]

Substituting the values we found for \( x - y \):

\[ ( \log 3 - \log 2 ) \cdot \frac{(\log_2(3) + 1)(\log_2(3) - 1)}{(\log_2(3))^2 - 1} \]

This leads us to the final answer.

### Final Answer

The final value of \( ( \log 3 - \log 2 )( x - y ) \) simplifies to \( 1 \).