If `(logx)/(b-c)=(logy)/(c-a)=(logz)/(a-b)` , then which of the following is/are true? `z y z=1` (b) `x^a y^b z^c=1` `x^(b+c)y^(c+b)=1` (d) `x y z=x^a y^b z^c`

Options

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

Explanation

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

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

where \( p \) is a constant. From this, we can express \( \log x \), \( \log y \), and \( \log z \) in terms of \( p \):

1. **Expressing \( \log x \), \( \log y \), and \( \log z \)**:
\[ \log x = p(b - c) \]
\[ \log y = p(c - a) \]
\[ \log z = p(a - b) \]

2. **Finding \( x \), \( y \), and \( z \)**:
We can convert the logarithmic expressions back to their exponential forms:
\[ x = k^{p(b - c)} \]
\[ y = k^{p(c - a)} \]
\[ z = k^{p(a - b)} \]

3. **Finding \( xyz \)**:
Now we multiply \( x \), \( y \), and \( z \):
\[ xyz = k^{p(b - c)} \cdot k^{p(c - a)} \cdot k^{p(a - b)} \]
Using the property of exponents (i.e., \( a^m \cdot a^n = a^{m+n} \)):
\[ xyz = k^{p(b - c + c - a + a - b)} = k^{p(0)} = k^0 = 1 \]
Thus, the first option \( xyz = 1 \) is true.

4. **Finding \( x^a y^b z^c \)**:
Now we check the second option:
\[ x^a y^b z^c = (k^{p(b - c)})^a \cdot (k^{p(c - a)})^b \cdot (k^{p(a - b)})^c \]
This simplifies to:
\[ = k^{p(a(b - c) + b(c - a) + c(a - b))} \]
Now we simplify the exponent:
\[ ab - ac + bc - ab + ac - bc = 0 \]
Therefore,
\[ x^a y^b z^c = k^0 = 1 \]
Hence, the second option \( x^a y^b z^c = 1 \) is also true.

5. **Finding \( x^{b+c} y^{c+b} \)**:
For the third option:
\[ x^{b+c} y^{c+b} = (k^{p(b - c)})^{b+c} \cdot (k^{p(c - a)})^{c+b} \]
This simplifies to:
\[ = k^{p(b+c)(b - c) + p(c+b)(c - a)} \]
Simplifying the exponent:
\[ = k^{p((b+c)(b-c) + (c+b)(c-a))} \]
After expanding and simplifying, we find that the exponent also sums to 0:
\[ = k^0 = 1 \]
Hence, the third option \( x^{b+c} y^{c+b} = 1 \) is true.

6. **Finding \( xyz = x^a y^b z^c \)**:
We already established that \( xyz = 1 \) and \( x^a y^b z^c = 1 \). Therefore, the fourth option \( xyz = x^a y^b z^c \) is also true.

**Final Conclusion**:
All options (a), (b), (c), and (d) are true.