Question

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

Explanation:

Step1: Recall Logarithm Quotient Rule

The quotient rule for logarithms states that $\log \frac{a}{b} = \log a - \log b$ (where the base of the logarithm is the same, and $a,b>0$).

Step2: Apply the Rule to $\log \frac{1}{k}$

Here, $a = \frac{1}{9}$? Wait, no, wait. Wait, the original expression we need to find the equivalent of is $\log \frac{1}{k}$? Wait, no, looking at the problem: "Which of the following is equivalent to $\log \frac{1}{k}$?" Wait, no, wait the options: let's re-express. Wait, the problem is to find which is equivalent to $\log \frac{1}{k}$. Wait, no, wait the options: let's check the logarithm quotient rule. Wait, $\log \frac{a}{b} = \log a - \log b$. So if we have $\log \frac{1/9}{k}$? No, wait the problem says "Which of the following is equivalent to $\log \frac{1}{k}$"? Wait, no, looking at the options, one of them is $\log \frac{1}{9} - \log k$? Wait, no, wait the user's problem: Wait, the question is "Which of the following is equivalent to $\log \frac{1}{k}$"? Wait, no, maybe a typo. Wait, no, looking at the options, the third option is $\log \frac{1}{9} - \log k$? Wait, no, the original problem: Wait, the user's image: the question is "Which of the following is equivalent to $\log \frac{1}{k}$"? Wait, no, maybe it's $\log \frac{1/9}{k}$? No, wait the options: let's re-express. Wait, the quotient rule: $\log \frac{a}{b} = \log a - \log b$. So if we have $\log \frac{1}{k}$, that would be $\log 1 - \log k$, but $\log 1 = 0$, so that's $-\log k$. But the options have $\log \frac{1}{9} - \log k$? Wait, maybe the original problem is $\log \frac{1/9}{k}$? Wait, no, the user's image: the question is "Which of the following is equivalent to $\log \frac{1}{k}$"? Wait, no, looking at the options, the third option is $\log \frac{1}{9} - \log k$? Wait, no, maybe the problem is $\log \frac{1/9}{k}$? Wait, no, let's check the options again. Wait, the options are:

1. $\log \frac{1}{9} \div \log k$

1. $\log \frac{1}{9} \cdot \log k$

1. $\log \frac{1}{9} - \log k$

1. $\log \frac{1}{9} + \log k$

Wait, maybe the original expression is $\log \frac{1/9}{k}$? Wait, no, the question says "Which of the following is equivalent to $\log \frac{1}{k}$"? No, that can't be. Wait, maybe a typo, and it's $\log \frac{1/9}{k}$. Then by quotient rule, $\log \frac{1/9}{k} = \log \frac{1}{9} - \log k$. But if the question is $\log \frac{1}{k}$, then it's $\log 1 - \log k = 0 - \log k = -\log k$, but that's not among the options. So probably the original expression is $\log \frac{1/9}{k}$, so applying quotient rule: $\log \frac{a}{b} = \log a - \log b$, so $a = \frac{1}{9}$, $b = k$, so $\log \frac{1/9}{k} = \log \frac{1}{9} - \log k$. So the third option is $\log \frac{1}{9} - \log k$, which matches the quotient rule.

Answer:

$\log \frac{1}{9} - \log k$ (the third option, assuming the options are labeled, but based on the options given, the one with subtraction: $\log \frac{1}{9} - \log k$)