If `log x/(b-c) =logy/(c-a) = logz/((a-b)` then the value of xyz is

Options

A. 1
B. 3
C. 2
D. none of these

Answer

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

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

Let's denote this common value as \( k \). Therefore, we can write:

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

Next, we can express \( x \), \( y \), and \( z \) in terms of \( k \):

\[ x = e^{k(b - c)} \]
\[ y = e^{k(c - a)} \]
\[ z = e^{k(a - b)} \]

Now, we need to find the product \( xyz \):

\[ xyz = e^{k(b - c)} \cdot e^{k(c - a)} \cdot e^{k(a - b)} \]

Using the property of exponents that states \( e^a \cdot e^b = e^{a + b} \), we can combine the exponents:

\[ xyz = e^{k[(b - c) + (c - a) + (a - b)]} \]

Now, simplify the expression inside the brackets:

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

Thus, we have:

\[ xyz = e^{k \cdot 0} = e^0 = 1 \]

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

\[ \boxed{1} \]