If `("log"3)/(x-y) = ("log"5)/(y-z) = ("log" 7)/(z-x), " then " 3^(x+y) 5^(y+z) 7^(z+x) =`

Options

A. 0
B. 2
C. 1
D. none of these

Answer

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

\[ \frac{\log 3}{x - y} = \frac{\log 5}{y - z} = \frac{\log 7}{z - x} = k \]

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

From the equation, we can express each logarithm in terms of \( k \):

\[ \log 3 = k(x - y) \]
\[ \log 5 = k(y - z) \]
\[ \log 7 = k(z - x) \]

### Step 2: Exponentiate to eliminate logarithms

Using the property \( \log a = b \implies a = 10^b \), we can rewrite the logarithmic equations in exponential form:

\[ 3 = 10^{k(x - y)} \]
\[ 5 = 10^{k(y - z)} \]
\[ 7 = 10^{k(z - x)} \]

### Step 3: Substitute into the expression

We need to evaluate the expression:

\[ 3^{x+y} \cdot 5^{y+z} \cdot 7^{z+x} \]

Substituting the values of \( 3 \), \( 5 \), and \( 7 \) from the previous steps:

\[ 3^{x+y} = (10^{k(x - y)})^{x+y} = 10^{k(x - y)(x + y)} \]
\[ 5^{y+z} = (10^{k(y - z)})^{y+z} = 10^{k(y - z)(y + z)} \]
\[ 7^{z+x} = (10^{k(z - x)})^{z+x} = 10^{k(z - x)(z + x)} \]

### Step 4: Combine the expressions

Now, we combine these expressions:

\[ 3^{x+y} \cdot 5^{y+z} \cdot 7^{z+x} = 10^{k(x - y)(x + y)} \cdot 10^{k(y - z)(y + z)} \cdot 10^{k(z - x)(z + x)} \]

Using the property of exponents \( a^m \cdot a^n = a^{m+n} \):

\[ = 10^{k\left((x - y)(x + y) + (y - z)(y + z) + (z - x)(z + x)\right)} \]

### Step 5: Simplify the exponent

Now, we simplify the exponent:

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

Combining these:

\[ x^2 - y^2 + y^2 - z^2 + z^2 - x^2 = 0 \]

### Step 6: Final result

Thus, the exponent simplifies to \( 0 \):

\[ 3^{x+y} \cdot 5^{y+z} \cdot 7^{z+x} = 10^{k \cdot 0} = 10^0 = 1 \]

Therefore, the final answer is:

\[ \boxed{1} \]