If `(logx)/(2a+3b-5c)=(log y)/(2b+3c-5a)=(log z)/(2c=3a-5b')` then xyz=

Options

A. 2
B. 1
C. 0
D. `-1`

Explanation

To solve the equation given in the question, we start with the expression:

\[ \frac{\log x}{2a + 3b - 5c} = \frac{\log y}{2b + 3c - 5a} = \frac{\log z}{2c + 3a - 5b} = k \]

### Step 1: Express each logarithm in terms of \( k \)

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

1. \(\log x = k(2a + 3b - 5c)\) (Equation 1)
2. \(\log y = k(2b + 3c - 5a)\) (Equation 2)
3. \(\log z = k(2c + 3a - 5b)\) (Equation 3)

### Step 2: Add the three logarithmic equations

Now, we will add these three equations:

\[ \log x + \log y + \log z = k(2a + 3b - 5c) + k(2b + 3c - 5a) + k(2c + 3a - 5b) \]

### Step 3: Factor out \( k \)

Factoring \( k \) out from the right-hand side gives us:

\[ \log x + \log y + \log z = k \left( (2a - 5a + 3a) + (3b + 2b - 5b) + (3c + 2c - 5c) \right) \]

### Step 4: Simplify the expression inside the parentheses

Now simplifying the terms inside the parentheses:

- For \( a \): \( 2a + 3a - 5a = 0 \)
- For \( b \): \( 3b + 2b - 5b = 0 \)
- For \( c \): \( 3c + 2c - 5c = 0 \)

Thus, we have:

\[ \log x + \log y + \log z = k \cdot 0 = 0 \]

### Step 5: Use the property of logarithms

Using the property of logarithms that states \(\log a + \log b + \log c = \log(abc)\), we can write:

\[ \log xyz = 0 \]

### Step 6: Solve for \( xyz \)

Taking the antilogarithm of both sides gives:

\[ xyz = e^0 = 1 \]

### Final Answer

Thus, the value of \( xyz \) is:

\[ \boxed{1} \]