Question

Using the Quotient Rule of Logarithms, the Product Rule of Logarithms, and the Power Rule of Logarithms, how can ln(13x2)
ln
⁡
(
1
3
𝑥
2
)
be fully expanded? (1 point)
Responses

−ln3−lnx
−
ln
⁡
3
−
ln
⁡
𝑥
negative l n 3 minus l n x

ln1−ln3−2lnx
ln
⁡
1
−
ln
⁡
3
−
2
ln
⁡
𝑥
l n 1 minus l n 3 minus 2 l n x

−ln3−2lnx
−
ln
⁡
3
−
2
ln
⁡
𝑥
negative l n 3 minus 2 l n x

−ln3−lnx2

Answer 1

To expand \( \ln\left(\frac{1}{3} x^2\right) \) using the Quotient Rule, Product Rule, and Power Rule of logarithms, we'll proceed as follows:

1. **Quotient Rule:** The Quotient Rule of logarithms states that \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \). For \( \ln\left(\frac{1}{3} x^2\right) \), we can separate it into \( \ln\left(\frac{1}{3}\right) + \ln(x^2) \).

Thus:
\[
\ln\left(\frac{1}{3} x^2\right) = \ln\left(\frac{1}{3}\right) + \ln(x^2)
\]

2. **Expanding \( \ln\left(\frac{1}{3}\right) \):** Using the property that \( \ln\left(\frac{1}{3}\right) = \ln(1) - \ln(3) \), and knowing that \( \ln(1) = 0 \), we have:
\[
\ln\left(\frac{1}{3}\right) = 0 - \ln(3) = -\ln(3)
\]

3. **Power Rule:** The Power Rule of logarithms states that \( \ln(a^b) = b \ln(a) \). Therefore, \( \ln(x^2) \) can be simplified as:
\[
\ln(x^2) = 2 \ln(x)
\]

4. **Combining everything:** Now we can combine all parts together:
\[
\ln\left(\frac{1}{3} x^2\right) = -\ln(3) + 2\ln(x)
\]

So, the fully expanded form of \( \ln\left(\frac{1}{3} x^2\right) \) is:
\[
-\ln(3) + 2\ln(x)
\]

Thus, the correct choice from the responses provided is:

**−ln3−2lnx** (negative l n 3 minus 2 l n x).