If `(logx)/(b-c)=(logy)/(c-a)=(logz)/(a-b)`. Then which of the following is true

Options

A. `xyz=1`
B. `x^(a)y^(b)z^(c )=1`
C. `x^(b+c)y^(c+a)z^(a+b)=1`
D. all are true

Answer

To solve the problem given the equation:

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

Let's denote the common value of these fractions as \( t \). Thus, we can express the logarithms of \( x \), \( y \), and \( z \) in terms of \( t \):

1. **Step 1: Express logarithms in terms of \( t \)**

\[ \log x = (b-c)t \]
\[ \log y = (c-a)t \]
\[ \log z = (a-b)t \]

**Hint:** Set the common ratio equal to a variable to simplify the expressions.

2. **Step 2: Add the logarithmic equations**

Now, we add all three equations:

\[ \log x + \log y + \log z = (b-c)t + (c-a)t + (a-b)t \]

The right-hand side simplifies as follows:

\[ (b-c + c-a + a-b)t = 0 \]

Therefore, we have:

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

**Hint:** Remember that the sum of logarithms can be expressed as the logarithm of the product.

3. **Step 3: Convert the sum of logarithms to a product**

Using the property of logarithms:

\[ \log(xyz) = 0 \]

This implies:

\[ xyz = 1 \]

**Hint:** Recall that if the logarithm of a number is zero, the number itself must equal one.

4. **Step 4: Analyze the second condition**

Now, let's analyze the second condition by multiplying the logarithmic equations by \( a \), \( b \), and \( c \) respectively:

\[ a \log x = a(b-c)t \]
\[ b \log y = b(c-a)t \]
\[ c \log z = c(a-b)t \]

Adding these gives:

\[ a \log x + b \log y + c \log z = (ab - ac + bc - ba + ca - cb)t = 0 \]

This leads to:

\[ \log(x^a y^b z^c) = 0 \]

Hence:

\[ x^a y^b z^c = 1 \]

**Hint:** Again, use the property of logarithms to convert the sum into a product.

5. **Step 5: Analyze the third condition**

For the third condition, we multiply the logarithmic equations by \( b+c \), \( c+a \), and \( a+b \):

\[ (b+c) \log x + (c+a) \log y + (a+b) \log z = 0 \]

This can be rearranged to:

\[ \log(x^{b+c} y^{c+a} z^{a+b}) = 0 \]

Thus:

\[ x^{b+c} y^{c+a} z^{a+b} = 1 \]

**Hint:** Again, use the logarithmic property to express the sum as a product.

6. **Conclusion: Final results**

From the steps above, we have derived three equations:

1. \( xyz = 1 \)
2. \( x^a y^b z^c = 1 \)
3. \( x^{b+c} y^{c+a} z^{a+b} = 1 \)

Therefore, all the options given in the problem are true.

**Final Answer:** All options are true.