Question

precal question: please check my answer!

Answer 1

I think it is choice D

Answer 2

You are right.

Answer 3

how do you know? @HolsterEmission

Answer 4

\[\frac{1}{2}\ln x+3\left[\ln(x-1)-\frac{2}{9}\ln(x+1)\right]\]Distribute the \(3\) to get
\[\frac{1}{2}\ln x+3\ln(x-1)-\frac{2}{3}\ln(x+1)\]Now use the power property for logs:
\[\ln x^{1/2}+\ln(x-1)^3+\ln(x+1)^{-2/3}\]with the last term rewritten again as
\[(x+1)^{-2/3}=\frac{1}{(x+1)^{2/3}}\]Fractional exponents can be rewritten as roots, so you have
\[\ln \sqrt x+\ln(x-1)^3+\ln\frac{1}{\sqrt[3]{(x+1)^2}}\]Finally, combine the terms using the property that sums of logs are equivalent to the log of a product, so you end up with
\[\ln\frac{\sqrt x(x-1)^3}{\sqrt[3]{(x+1)^2}}\]