If a, b, c are positive real numbers, then
`(1)/("log"_(ab)abc) + (1)/("log"_(bc)abc) + (1)/("log"_(ca)abc) =`

Options

A. 0
B. 1
C. 2
D. none of these

Explanation

To solve the expression

\[ \frac{1}{\log_{ab} abc} + \frac{1}{\log_{bc} abc} + \frac{1}{\log_{ca} abc} \]

we will follow these steps:

### Step 1: Rewrite the logarithmic terms
Using the property of logarithms that states \( \log_a b = \frac{1}{\log_b a} \), we can rewrite each term:

\[ \frac{1}{\log_{ab} abc} = \log_{abc} (ab) \]
\[ \frac{1}{\log_{bc} abc} = \log_{abc} (bc) \]
\[ \frac{1}{\log_{ca} abc} = \log_{abc} (ca) \]

Thus, the expression becomes:

\[ \log_{abc} (ab) + \log_{abc} (bc) + \log_{abc} (ca) \]

### Step 2: Combine the logarithmic terms
Using the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \), we can combine the logarithmic terms:

\[ \log_{abc} (ab) + \log_{abc} (bc) + \log_{abc} (ca) = \log_{abc} (ab \cdot bc \cdot ca) \]

### Step 3: Simplify the argument of the logarithm
Now, we simplify the product:

\[ ab \cdot bc \cdot ca = a^2b^2c^2 \]

Thus, we have:

\[ \log_{abc} (a^2b^2c^2) \]

### Step 4: Apply the power property of logarithms
Using the property that \( \log_a (b^c) = c \cdot \log_a b \), we can simplify further:

\[ \log_{abc} (a^2b^2c^2) = 2 \cdot \log_{abc} (abc) \]

### Step 5: Evaluate the logarithm
Since \( \log_{abc} (abc) = 1 \) (because the base and the argument are the same), we find:

\[ 2 \cdot \log_{abc} (abc) = 2 \cdot 1 = 2 \]

### Final Answer
Thus, the value of the expression is:

\[ \boxed{2} \]

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