Solve: Complex Logarithmic Fraction with log₄7 and log₄₉a Terms

$\frac{\log_47\times\log_{\frac{1}{49}}a}{c\log_4b}=$

To solve this problem, we'll follow these steps:

• Step 1: Express $\log_4{7}$ and $\log_{\frac{1}{49}}{a}$ using the change-of-base formula.
• Step 2: Simplify the product $\log_4{7} \times \log_{\frac{1}{49}}{a}$.
• Step 3: Simplify the entire expression by using logarithmic identities.

Let's work through each step:
Step 1: Using the change-of-base formula, $\log_4{7} = \frac{\log_k{7}}{\log_k{4}}$ and $\log_{\frac{1}{49}}{a} = \frac{\log_k{a}}{\log_k{\frac{1}{49}}}$. Choose $k = 10$ (common log) for simplicity.
Note that $\log_k{\frac{1}{49}} = \log_k{49^{-1}} = -\log_k{49}$. Also, $49 = 7^2$, so $\log_k{49} = 2\log_k{7}$. Therefore, $\log_{\frac{1}{49}}{a} = \frac{\log_k{a}}{-2\log_k{7}}$.

Step 2: The product $\log_4{7} \times \log_{\frac{1}{49}}{a} = \left(\frac{\log_k{7}}{\log_k{4}}\right)\left(\frac{\log_k{a}}{-2\log_k{7}}\right)$ simplifies to $\frac{\log_k{a}}{-2\log_k{4}}$ after canceling $\log_k{7}$.

Step 3: The expression becomes $\frac{\frac{\log_k{a}}{-2\log_k{4}}}{c\log_4{b}}$, which simplifies to $\frac{\log_k{a}}{-2c\log_k{4}\log_4{b}}$. Convert $\log_4{b}$ into $\frac{\log_k{b}}{\log_k{4}}$, leading to $\frac{\log_k{a}}{-2c\log_k{b}}$. Using the change-of-base formula again, this gives $-\frac{1}{2}\log_{b^c}{a}$.

This can be rewritten using inverse log properties as $\log_{b^c}{\left(\frac{1}{\sqrt{a}}\right)}$.

Therefore, the solution to the problem is $\log_{b^c}\frac{1}{\sqrt{a}}$.