The value of `("log"_(a)("log"_(b)a))/("log"_(b)("log"_(a)b))`, is

Options

A. `"log"_(b)a`
B. `"log"_(a) b`
C. `-"log"_(a)b`
D. `-"log"_(b)a`

Answer

To solve the expression \(\frac{\log_a(\log_b a)}{\log_b(\log_a b)}\), we will use the change of base formula for logarithms, which states that \(\log_a b = \frac{\log b}{\log a}\).

### Step-by-Step Solution:

1. **Rewrite the logarithms using the change of base formula:**
\[ \log_a(\log_b a) = \frac{\log(\log_b a)}{\log a} \]
\[ \log_b(\log_a b) = \frac{\log(\log_a b)}{\log b} \]

2. **Substituting these into the original expression:**
\[ \frac{\log_a(\log_b a)}{\log_b(\log_a b)} = \frac{\frac{\log(\log_b a)}{\log a}}{\frac{\log(\log_a b)}{\log b}} = \frac{\log(\log_b a)}{\log(\log_a b)} \cdot \frac{\log b}{\log a} \]

3. **Now, we need to simplify \(\log(\log_b a)\) and \(\log(\log_a b)\):**
- Using the change of base formula again:
\[ \log_b a = \frac{\log a}{\log b} \implies \log(\log_b a) = \log\left(\frac{\log a}{\log b}\right) = \log(\log a) - \log(\log b) \]
\[ \log_a b = \frac{\log b}{\log a} \implies \log(\log_a b) = \log\left(\frac{\log b}{\log a}\right) = \log(\log b) - \log(\log a) \]

4. **Substituting these back into the expression:**
\[ \frac{\log(\log_b a)}{\log(\log_a b)} = \frac{\log(\log a) - \log(\log b)}{\log(\log b) - \log(\log a)} \]

5. **Now, substitute this back into our expression:**
\[ \frac{\log(\log a) - \log(\log b)}{\log(\log b) - \log(\log a)} \cdot \frac{\log b}{\log a} \]

6. **Notice that \(\log(\log b) - \log(\log a) = -(\log(\log a) - \log(\log b))\):**
\[ = -1 \cdot \frac{\log b}{\log a} = -\frac{\log b}{\log a} \]

7. **Thus, the final value of the expression is:**
\[ -\frac{\log b}{\log a} = -\log_a b \]

### Final Answer:
The value of \(\frac{\log_a(\log_b a)}{\log_b(\log_a b)}\) is \(-\log_a b\).