Certainly! Let's analyze the logarithmic expression [tex]\(\log \left(\frac{1}{n}\right)\)[/tex].
We can use the properties of logarithms to simplify this expression. One important property of logarithms that is useful here is:
[tex]\[
\log \left(\frac{a}{b}\right) = \log a - \log b
\][/tex]
In this case, we can apply this property with [tex]\(a = 1\)[/tex] and [tex]\(b = n\)[/tex]:
[tex]\[
\log \left(\frac{1}{n}\right) = \log 1 - \log n
\][/tex]
Next, we need to evaluate [tex]\(\log 1\)[/tex]. It is a well-known fact that the logarithm of 1 in any base is always 0:
[tex]\[
\log 1 = 0
\][/tex]
So we substitute [tex]\(\log 1\)[/tex] with 0 in the expression:
[tex]\[
\log \left(\frac{1}{n}\right) = 0 - \log n
\][/tex]
Simplifying this, we get:
[tex]\[
\log \left(\frac{1}{n}\right) = -\log n
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{-\log n}
\][/tex]
So the answer to the question is option (d) [tex]\(-\log n\)[/tex].
