`log (log_(ab) a +(1)/(log_(b)ab)` is (where `ab ne 1)`

Options

A. 0
B. 1
C. `log_(a)ab`
D. `log_(b)ab`

Answer

To solve the problem \( \log\left(\log_{ab} a + \frac{1}{\log_b ab}\right) \) where \( ab \neq 1 \), we will break it down step by step.

### Step 1: Rewrite the logarithms using properties
We start by rewriting the logarithms using the change of base formula. The change of base formula states that:
\[ \log_b a = \frac{1}{\log_a b} \]
Thus, we can rewrite \( \log_{ab} a \) and \( \log_b ab \).

1. **Rewrite \( \log_{ab} a \)**:
\[ \log_{ab} a = \frac{\log a}{\log(ab)} = \frac{\log a}{\log a + \log b} \]

2. **Rewrite \( \log_b ab \)**:
\[ \log_b ab = \frac{\log(ab)}{\log b} = \frac{\log a + \log b}{\log b} \]

### Step 2: Substitute back into the expression
Now substitute these back into the original expression:
\[ \log\left(\frac{\log a}{\log a + \log b} + \frac{1}{\frac{\log a + \log b}{\log b}}\right) \]

### Step 3: Simplify the expression
Now simplify the expression inside the logarithm:
1. The second term simplifies as follows:
\[ \frac{1}{\frac{\log a + \log b}{\log b}} = \frac{\log b}{\log a + \log b} \]

2. Now combine the two terms:
\[ \frac{\log a}{\log a + \log b} + \frac{\log b}{\log a + \log b} = \frac{\log a + \log b}{\log a + \log b} = 1 \]

### Step 4: Final logarithm
Now we have:
\[ \log(1) \]
Since the logarithm of 1 is always 0, we conclude that:
\[ \log(1) = 0 \]

### Final Answer
Thus, the answer to the question is:
\[ \boxed{0} \]

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