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

Options

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

Answer

To solve the equation \( (11)^a = (19)^b = (209)^c \), we can set all three expressions equal to a common variable \( K \). Thus, we have:

1. \( 11^a = K \)
2. \( 19^b = K \)
3. \( 209^c = K \)

From these equations, we can express \( 11 \), \( 19 \), and \( 209 \) in terms of \( K \):

\[ 11 = K^{1/a} \]
\[ 19 = K^{1/b} \]
\[ 209 = K^{1/c} \]

Next, we can multiply the first two equations:

\[ 11 \times 19 = K^{1/a} \times K^{1/b} = K^{(1/a) + (1/b)} \]

Now, we know that \( 11 \times 19 = 209 \):

\[ 209 = K^{(1/a) + (1/b)} \]

Since we also have \( 209 = K^{1/c} \), we can set these equal to each other:

\[ K^{(1/a) + (1/b)} = K^{1/c} \]

Since the bases are the same, we can equate the exponents:

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

Now, let's multiply through by \( abc \) to eliminate the denominators:

\[ bc + ac = ab \]

Rearranging gives us:

\[ ab = ac + bc \]

This can be rewritten as:

\[ ab - ac - bc = 0 \]

Thus, we have derived the relationship between \( a \), \( b \), and \( c \):

\[ ab = ac + bc \]