`1/log_(ab)(abc)+1/log_(bc)(abc)+1/log_(ca)(abc)` is equal to:

Options

A. `1//2`
B. 1
C. 2
D. 4

Explanation

To solve the expression \( \frac{1}{\log_{ab}(abc)} + \frac{1}{\log_{bc}(abc)} + \frac{1}{\log_{ca}(abc)} \), we can use the change of base formula for logarithms, which states that:

\[ \log_a b = \frac{\log_c b}{\log_c a} \]

### Step 1: Apply the Change of Base Formula
Using the change of base formula, we can rewrite each term in the expression:

\[ \log_{ab}(abc) = \frac{\log(abc)}{\log(ab)} = \frac{\log(abc)}{\log(a) + \log(b)} \]

\[ \log_{bc}(abc) = \frac{\log(abc)}{\log(b) + \log(c)} \]

\[ \log_{ca}(abc) = \frac{\log(abc)}{\log(c) + \log(a)} \]

### Step 2: Rewrite the Expression
Now substituting these into the original expression, we have:

\[ \frac{1}{\log_{ab}(abc)} = \frac{\log(ab)}{\log(abc)} = \frac{\log(a) + \log(b)}{\log(abc)} \]

\[ \frac{1}{\log_{bc}(abc)} = \frac{\log(b) + \log(c)}{\log(abc)} \]

\[ \frac{1}{\log_{ca}(abc)} = \frac{\log(c) + \log(a)}{\log(abc)} \]

Thus, the entire expression becomes:

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

### Step 3: Combine the Fractions
Now, we can combine these fractions since they all have the same denominator:

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

### Step 4: Simplify the Numerator
In the numerator, we can combine like terms:

\[ = \frac{2\log(a) + 2\log(b) + 2\log(c)}{\log(abc)} \]

### Step 5: Factor Out the Common Terms
Factoring out the common factor of 2:

\[ = \frac{2(\log(a) + \log(b) + \log(c))}{\log(abc)} \]

### Step 6: Use the Property of Logarithms
Using the property of logarithms that states \( \log(abc) = \log(a) + \log(b) + \log(c) \):

\[ = \frac{2\log(abc)}{\log(abc)} \]

### Step 7: Final Simplification
Now, we can simplify this expression:

\[ = 2 \]

Thus, the final answer is:

\[ \boxed{2} \]