If `(logx)/(b-c)=(logy)/(c-a)=(logz)/(a-b)`, then `x^(a).y^(b).z^(c)` =

Options

A. a
B. abc
C. xyz
D. none of these

Answer

To solve the equation \( \frac{\log x}{b-c} = \frac{\log y}{c-a} = \frac{\log z}{a-b} \), we can denote the common ratio as \( t \). This gives us three equations:

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

Now, we will add these three equations together:

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

On simplifying the right-hand side, we see that:

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

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:

\[ xyz = 1 \]

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

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

Adding these equations gives:

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

Now, simplifying the right-hand side:

\[ at(b - c) + bt(c - a) + ct(a - b) = abt - act + bct - abt + cat - bct = 0 \]

Thus, we have:

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

Using the property of logarithms again, we can write:

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

This implies:

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

Therefore, the final result is:

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