Choose the expression that shows an expanded form of the logarithm. $$ \begin{array}{l}\ln \left(\frac{x^{20} \sqrt{x-1}}{3 x-11}\right) \ 20 \ln x+\ln \sqrt{x-1}-\ln (3 x-12) \ \left(\frac{1}{2}\right) \ln x^{20}(x-1)-\ln (3 x-11) \ 20 \ln x+\left(\frac{1}{2}\right) \ln (x-1)-\ln (3 x-11)\end{array} $$

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UpStudy AI · Accepted Answer

1. Start with the expression:
   $$
   \ln \left(\frac{x^{20} \sqrt{x-1}}{3x-11}\right)
   $$

2. Use the logarithm quotient rule:
   $$
   \ln \left(\frac{A}{B}\right)=\ln A-\ln B
   $$
   so we have:
   $$
   \ln \left(\frac{x^{20} \sqrt{x-1}}{3x-11}\right)=\ln \left(x^{20} \sqrt{x-1}\right)-\ln (3x-11)
   $$

3. Now apply the logarithm product rule to $$\ln \left(x^{20} \sqrt{x-1}\right)$$:
   $$
   \ln (AB)=\ln A + \ln B
   $$
   which gives:
   $$
   \ln \left(x^{20} \sqrt{x-1}\right)=\ln (x^{20})+\ln \sqrt{x-1}
   $$

4. Use the power rule for logarithms: $$\ln (x^n)=n \ln x$$. For the first part:
   $$
   \ln (x^{20})=20 \ln x
   $$

5. Recognize that $$\sqrt{x-1}=(x-1)^{\frac{1}{2}}$$, so:
   $$
   \ln \sqrt{x-1}=\frac{1}{2} \ln (x-1)
   $$

6. Putting it all together, we obtain the fully expanded form:
   $$
   20 \ln x + \frac{1}{2}\ln (x-1) - \ln (3x-11)
   $$

Thus, the correct expanded form is:
$$
20 \ln x+\frac{1}{2}\ln (x-1)-\ln (3x-11)
$$
The expanded form is $$20 \ln x + \frac{1}{2} \ln (x-1) - \ln (3x-11)$$.

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