If `(11)^a = (19)^b = (209)^c` then the correct relation is :

Options

A. `ab = (c)^(ab)`
B. `a^b = c`
C. `ab = c(a + b)`
D. None of these

Answer

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

\[ (11)^a = (19)^b = (209)^c = k \]

where \( k \) is some constant.

### Step 1: Express each base in terms of \( k \)

From the equation, we can express each base in terms of \( k \):

1. For \( 11^a = k \):
\[ k = 11^a \implies k^{\frac{1}{a}} = 11 \]

2. For \( 19^b = k \):
\[ k = 19^b \implies k^{\frac{1}{b}} = 19 \]

3. For \( 209^c = k \):
\[ k = 209^c \implies k^{\frac{1}{c}} = 209 \]

### Step 2: Relate the bases using the multiplication property

We know that \( 11 \times 19 = 209 \). Therefore, we can write:

\[ 11 \cdot 19 = 209 \]

Substituting the expressions we found for \( k \):

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

### Step 3: Combine the exponents

Using the property of exponents, we can combine the left side:

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

### Step 4: Set the exponents equal to each other

Since the bases are the same, we can set the exponents equal to each other:

\[ \frac{1}{a} + \frac{1}{b} = \frac{1}{c} \]

### Step 5: Rearranging the equation

To find a relation among \( a \), \( b \), and \( c \), we can rearrange the equation:

\[ \frac{1}{c} = \frac{1}{a} + \frac{1}{b} \]

This can be rewritten as:

\[ \frac{1}{c} = \frac{b + a}{ab} \]

Thus, we can express \( c \) in terms of \( a \) and \( b \):

\[ c = \frac{ab}{a + b} \]

### Final Relation

From the above steps, we conclude that the correct relation is:

\[ c = \frac{ab}{a + b} \]