If `(a(b + c - a))/(log a) = (b (c + a - b))/(log b) = (c (a + b - c))/(log c ) ` then `(a^(b).b^(a))/(c^(a).a^(c))` equals

Options

A. `(1)/(2) `
B. `- 1`
C. 1
D. none of these

Answer

To solve the equation

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

we will denote the common value of these expressions as \( k \).

### Step 1: Set up the equations

From the given equation, we can write:

1. \(\log a = \frac{a(b + c - a)}{k}\)
2. \(\log b = \frac{b(c + a - b)}{k}\)
3. \(\log c = \frac{c(a + b - c)}{k}\)

### Step 2: Express \( \log a^b \) and \( \log b^a \)

We can multiply the first equation by \( b \):

\[ b \log a = \frac{ab(b + c - a)}{k} \]

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

\[ \log a^b = \frac{ab(b + c - a)}{k} \]

Similarly, for \( \log b^a \):

\[ a \log b = \frac{ab(c + a - b)}{k} \]

Thus,

\[ \log b^a = \frac{ab(c + a - b)}{k} \]

### Step 3: Add the two logarithmic equations

Now, adding the two logarithmic equations:

\[ \log a^b + \log b^a = \frac{ab(b + c - a)}{k} + \frac{ab(c + a - b)}{k} \]

Factoring out \( \frac{ab}{k} \):

\[ \log (a^b \cdot b^a) = \frac{ab}{k} \left( (b + c - a) + (c + a - b) \right) \]

Simplifying the right side:

\[ = \frac{ab}{k} \cdot (2c) \]

### Step 4: Express \( \log c^a \) and \( \log a^c \)

Now, we will do the same for \( c \):

\[ a \log c = \frac{ac(a + b - c)}{k} \]
\[ \log c^a = \frac{ac(a + b - c)}{k} \]

And for \( a \):

\[ c \log a = \frac{c(b + c - a)}{k} \]
\[ \log a^c = \frac{c(b + c - a)}{k} \]

### Step 5: Add the two logarithmic equations for \( c \)

Adding these:

\[ \log (c^a \cdot a^c) = \frac{ac(a + b - c)}{k} + \frac{c(b + c - a)}{k} \]

Factoring out \( \frac{ac}{k} \):

\[ = \frac{ac}{k} \left( (a + b - c) + (b + c - a) \right) \]

Simplifying gives:

\[ = \frac{ac}{k} \cdot (2b) \]

### Step 6: Equate the two results

Now we have:

\[ \log (a^b \cdot b^a) = \frac{2abc}{k} \]
\[ \log (c^a \cdot a^c) = \frac{2abc}{k} \]

Since both expressions are equal, we can write:

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

### Step 7: Exponentiate both sides

Exponentiating both sides gives:

\[ a^b \cdot b^a = c^a \cdot a^c \]

### Step 8: Rearranging the equation

Rearranging gives:

\[ \frac{a^b \cdot b^a}{c^a \cdot a^c} = 1 \]

Thus, the final result is:

\[ \frac{a^b \cdot b^a}{c^a \cdot a^c} = 1 \]