If `(log x)/(l + m -2n) = (log y)/(m + n -2l) = (log z)/(n + l -2m)`, then xyz is equal to

Options

A. 0
B. 1
C. lmn
D. 2

Answer

To solve the equation

\[ \frac{\log x}{l + m - 2n} = \frac{\log y}{m + n - 2l} = \frac{\log z}{n + l - 2m} \]

we can introduce a common variable \( k \) such that:

\[ \frac{\log x}{l + m - 2n} = k, \quad \frac{\log y}{m + n - 2l} = k, \quad \frac{\log z}{n + l - 2m} = k \]

From these equations, we can express \( \log x \), \( \log y \), and \( \log z \) in terms of \( k \):

1. **For \( x \)**:
\[ \log x = k (l + m - 2n) \]

2. **For \( y \)**:
\[ \log y = k (m + n - 2l) \]

3. **For \( z \)**:
\[ \log z = k (n + l - 2m) \]

Next, we want to find the product \( xyz \):

\[ \log(xyz) = \log x + \log y + \log z \]

Substituting the expressions we found:

\[ \log(xyz) = k (l + m - 2n) + k (m + n - 2l) + k (n + l - 2m) \]

Factoring out \( k \):

\[ \log(xyz) = k \left( (l + m - 2n) + (m + n - 2l) + (n + l - 2m) \right) \]

Now, let's simplify the expression inside the parentheses:

\[ = k \left( l + m - 2n + m + n - 2l + n + l - 2m \right) \]

Combining like terms:

- For \( l \): \( l - 2l + l = 0 \)
- For \( m \): \( m - 2m + m = 0 \)
- For \( n \): \( -2n + n + n = 0 \)

Thus, we have:

\[ \log(xyz) = k \cdot 0 = 0 \]

This implies:

\[ xyz = e^0 = 1 \]

Therefore, the final result is:

\[ xyz = 1 \]