if `loga/(b-c)=logb/(c-a)=logc/(a-b)` then find the value of `a^ab^bc^c`

Options

A. 0
B. 1
C. abc
D. none of these

Answer

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

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

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

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

### Step 1: Express \(\log a\), \(\log b\), and \(\log c\) in terms of \( k \)

From the equations we have:

\[ \log a = k(b - c) \]
\[ \log b = k(c - a) \]
\[ \log c = k(a - b) \]

### Step 2: Add all three equations

Now, we add all three equations:

\[ \log a + \log b + \log c = k(b - c) + k(c - a) + k(a - b) \]

### Step 3: Simplify the right-hand side

The right-hand side simplifies as follows:

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

Thus, we have:

\[ \log a + \log b + \log c = 0 \]

### Step 4: Apply the property of logarithms

Using the property of logarithms that states \(\log x + \log y + \log z = \log(xyz)\), we can rewrite the left-hand side:

\[ \log(abc) = 0 \]

### Step 5: Exponentiate both sides

Exponentiating both sides gives us:

\[ abc = 10^0 = 1 \]

### Step 6: Find \( a^a b^b c^c \)

Now we need to find the value of \( a^a b^b c^c \). We already know:

\[ \log a + \log b + \log c = 0 \]

This implies:

\[ \log(a^a) + \log(b^b) + \log(c^c) = a \log a + b \log b + c \log c \]

Using the earlier derived equations:

\[ a \log a = a \cdot k(b - c) \]
\[ b \log b = b \cdot k(c - a) \]
\[ c \log c = c \cdot k(a - b) \]

Adding these gives:

\[ a \log a + b \log b + c \log c = k(ab - ac + bc - ba + ca - cb) = k(0) = 0 \]

Thus:

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

### Step 7: Exponentiate again

Exponentiating both sides gives us:

\[ a^a b^b c^c = 10^0 = 1 \]

### Final Answer

Therefore, the value of \( a^a b^b c^c \) is:

\[ \boxed{1} \]