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

Options

A. 0
B. 1
C. 2
D. 3

Answer

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. The change of base formula states that:

\[ \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \]

We can apply this property to each term in our expression.

### Step 1: Rewrite each logarithm using the change of base formula

Using the change of base formula, we can rewrite each term:

\[ \log_{ab} abc = \frac{\log abc}{\log ab} \]
\[ \log_{bc} abc = \frac{\log abc}{\log bc} \]
\[ \log_{ca} abc = \frac{\log abc}{\log ca} \]

### Step 2: Substitute back into the expression

Now, substituting these back into the original expression gives us:

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

Thus, we can rewrite the entire expression as:

\[ \frac{\log ab}{\log abc} + \frac{\log bc}{\log abc} + \frac{\log ca}{\log abc} \]

### Step 3: Combine the fractions

Since all terms have a common denominator, we can combine them:

\[ \frac{\log ab + \log bc + \log ca}{\log abc} \]

### Step 4: Use the property of logarithms

Using the property of logarithms that states \( \log a + \log b = \log(ab) \), we can simplify the numerator:

\[ \log ab + \log bc + \log ca = \log(a \cdot b) + \log(b \cdot c) + \log(c \cdot a) \]

This can be rewritten as:

\[ \log(abc \cdot ab \cdot bc \cdot ca) = \log(a^2 b^2 c^2) \]

### Step 5: Final simplification

Now substituting this back into our expression gives:

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

Using the property of logarithms again, we can simplify:

\[ \log(a^2 b^2 c^2) = 2 \log(abc) \]

So, we have:

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

### Final Answer

Thus, the final result is:

\[ \frac{1}{\log_{ab} abc} + \frac{1}{\log_{bc} abc} + \frac{1}{\log_{ca} abc} = 2 \]