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. `xyz=x^(a)y^(b)z^(c)`

Answer

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

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

From this, we can express \(\log x\), \(\log y\), and \(\log z\) in terms of \(t\):

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

Next, we will add these three equations together:

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

Now, simplifying the right-hand side:

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

Thus, we have:

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

Using the property of logarithms, we can rewrite this as:

\[ \log(xyz) = 0 \]

This implies:

\[ xyz = 1 \]

Now, we can explore the implications of this result.

Next, we multiply each of the original equations by \(a\), \(b\), and \(c\) respectively:

1. From Equation 1, multiplying by \(a\):
\[ a \log x = a t(b - c) \]
2. From Equation 2, multiplying by \(b\):
\[ b \log y = b t(c - a) \]
3. From Equation 3, multiplying by \(c\):
\[ c \log z = c t(a - b) \]

Adding these equations together gives:

\[ a \log x + b \log y + c \log z = at(b - c) + bt(c - a) + ct(a - b) \]

The right-hand side simplifies to:

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

Thus, we have:

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

Using the property of logarithms again, we can express this as:

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

This implies:

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

Now, let's summarize the results we have derived:

1. \(xyz = 1\)
2. \(x^a y^b z^c = 1\)

Both of these statements are true based on the original equations.