Correct Option: B To find the value of the given expression, we will simplify each term individually using the fundamental properties of logarithms. **Step 1: Simplify the term $\frac{1}{1+x}$** Given $x = \log_a (bc)$. First, let's evaluate $1+x$: $1+x = 1 + \log_a (bc)$ Using the logarithm property that $1 = \log_k k$ for any base $k$, we can write $1$ as $\log_a a$: $1+x = \log_a a + \log_a (bc)$ Now, apply the logarithm product rule, $\log_p m + \log_p n = \log_p (mn)$: $1+x = \log_a (a \cdot bc) = \log_a (abc)$ Substitute this back into the expression $\frac{1}{1+x}$: $\frac{1}{1+x} = \frac{1}{\log_a (abc)}$ Using the change of base property, $\frac{1}{\log_p q} = \log_q p$: $\frac{1}{1+x} = \log_{abc} a$ **Step 2: Simplify the term $\frac{1}{1+y}$** Given $y = \log_b (ac)$. Following the same approach as for $x$: $1+y = 1 + \log_b (ac)$ $1+y = \log_b b + \log_b (ac)$ $1+y = \log_b (b \cdot ac) = \log_b (abc)$ Therefore, applying the change of base property: $\frac{1}{1+y} = \frac{1}{\log_b (abc)} = \log_{abc} b$ **Step 3: Simplify the term $\frac{1}{1+z}$** Given $z = \log_c (ab)$. Similarly: $1+z = 1 + \log_c (ab)$ $1+z = \log_c c + \log_c (ab)$ $1+z = \log_c (c \cdot ab) = \log_c (abc)$ Therefore, applying the change of base property: $\frac{1}{1+z} = \frac{1}{\log_c (abc)} = \log_{abc} c$ **Step 4: Sum the simplified terms** Now, we need to find the value of the sum: $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} = \log_{abc} a + \log_{abc} b + \log_{abc} c$ Apply the logarithm product rule again for multiple terms, $\log_p m + \log_p n + \log_p o = \log_p (mno)$: Sum $= \log_{abc} (a \cdot b \cdot c)$ Sum $= \log_{abc} (abc)$ Finally, using the property $\log_k k = 1$: Sum $= 1$ Thus, the value of the expression is $1$. The final answer is $\boxed{1}$.