Question

Using numerical analysis and the Power Rule of Logarithms, logxk=klogx
log
⁡
𝑥
𝑘
=
𝑘
log
⁡
𝑥
, how can log(1100)
log
⁡
(
1
100
)
be rewritten?(1 point)
Responses

2log−10
2
log
−
10
2 logarithm negative 10

10log−2
10
log
−
2
10 logarithm negative 2

log−20
log
−
20
logarithm negative 20

−2log10

Answer 1

To simplify the expression \(\frac{\log(1100)}{\log(1/100)}\) using the Power Rule of Logarithms:

1. First, rewrite \(1/100\) as \(10^{-2}\). This gives us:
\[
\log(1/100) = \log(10^{-2})
\]

2. Now, applying the Power Rule of Logarithms:
\[
\log(10^{-2}) = -2 \log(10)
\]

3. Substitute this back into the original expression:
\[
\frac{\log(1100)}{\log(1/100)} = \frac{\log(1100)}{-2 \log(10)}
\]

4. This is equivalent to:
\[
-\frac{1}{2} \frac{\log(1100)}{\log(10)}
\]

Now, we can express this as:
\[
-\frac{1}{2} \cdot \log_{10}(1100)
\]

The given responses seem to refer to simplified forms rather than maintaining \(\log(1100)\) in the numerator. However, focusing on the expressions you've listed, the closest match after simplification is:

\[
-2 \log_{10}(2)
\]

This is incorrect as the derived relationship does not provide exactly this term unless further context is given regarding the original logarithm bases or values inside logarithms. But to resolve directly from \(\log(1100)\):

The answer \(-2 \log_{10}(2)\) seems to suggest a different pathway not directly derivable from given expressions. Ultimately:
\(\mathbf{\text{-2 log_{10}}}\) is the direct result after accounting for the negatives and logarithmic conversion.

So, based on this analysis:

\(\mathbf{-2 \log_{10}}\) is the one indicated from your multiple-choice responses fitting the equivalency derived.