If log x : log y : log z =(y - z): (z -x): (x - y), then

Options

A. `X^(y) Y^(z) Z^(x) = 1`
B. `X^(x) Y^(y) Z^(z) = 1`
C. `x sqrtx. y sqrty. z sqrtz = 1`
D. None of these

Answer

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

\[ \frac{\log x}{\log y} = \frac{y - z}{z - x} = \frac{z - x}{x - y} \]

### Step 1: Set the ratios equal to a constant \( k \)

We can express the ratios in terms of a constant \( k \):

\[ \frac{\log x}{y - z} = k \quad \text{and} \quad \frac{\log y}{z - x} = k \quad \text{and} \quad \frac{\log z}{x - y} = k \]

### Step 2: Rearranging the equations

From the first equation, we can express \( \log x \):

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

From the second equation, we express \( \log y \):

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

From the third equation, we express \( \log z \):

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

### Step 3: Summing the logarithmic expressions

Now, we can add these three equations:

\[ \log x + \log y + \log z = k(y - z) + k(z - x) + k(x - y) \]

### Step 4: Simplifying the right-hand side

Notice that the right-hand side simplifies:

\[ k(y - z + z - x + x - y) = k(0) = 0 \]

### Step 5: Concluding the logarithmic equation

Thus, we have:

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

This implies:

\[ \log(xyz) = 0 \]

### Step 6: Exponentiating both sides

Exponentiating both sides gives:

\[ xyz = 1 \]

### Step 7: Finding the expression for \( x^x y^y z^z \)

Now, we need to check the expression \( x^x y^y z^z \). We can express it using our earlier results:

Using the property of logarithms:

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

Substituting the values we found:

\[ x \log x = kx(y - z), \quad y \log y = ky(z - x), \quad z \log z = kz(x - y) \]

### Step 8: Summing these values

We can sum these:

\[ x \log x + y \log y + z \log z = k(x(y - z) + y(z - x) + z(x - y)) \]

The expression inside the parentheses simplifies to zero, as shown before.

### Final Result

Thus, we find:

\[ \log(x^x y^y z^z) = 0 \implies x^x y^y z^z = 1 \]

### Conclusion

The correct answer is:

\[ x^x y^y z^z = 1 \]