Question

which of the following is equivalent to \\(\log \frac{\frac{1}{9}}{k}\\)?
\\(\circ\\) \\(\log \frac{1}{9} - \log k\\)
\\(\circ\\) \\(\log \frac{1}{9} \div \log k\\)
\\(\circ\\) \\(\log \frac{1}{9} \cdot \log k\\)
\\(\circ\\) \\(\log \frac{1}{9} + \log k\\)

Explanation:

Step1: Recall the logarithm quotient rule

The quotient rule for logarithms states that $\log_b \frac{m}{n} = \log_b m - \log_b n$, where $b>0$, $b
eq1$, $m>0$, and $n>0$.

Step2: Apply the quotient rule to $\log \frac{\frac{1}{9}}{k}$

Here, $m = \frac{1}{9}$ and $n = k$, and the base of the logarithm (though not explicitly written, we assume a common base, e.g., base 10 or natural logarithm base $e$) follows the quotient rule. So, $\log \frac{\frac{1}{9}}{k} = \log \frac{1}{9} - \log k$.

Answer:

$\log \frac{1}{9} - \log k$ (the first option)