Express as a single logarithm. Simplify. $$ \frac{1}{3} \ln (x+2)^3+\frac{1}{2}\left[\ln x-\ln \left(x^2+3 x+2\right)^2\right] $$
So I am posting the question, how I solved it and then how the TA told me to solve it differently because he said you need to root of $x^2$ is $|x|$. However when my professor posted the solutions, my original answer was correct, so I am confused...
My Original Answer: $$ \frac{1}{3} \ln (x+2)^3+\frac{1}{2}\left[\ln x-\ln \left(x^2+3 x+2\right)^2\right] \\=\ln \left[(x+2)^3\right]^{1 / 3}+\frac{1}{2} \ln\left(\frac{x}{\left(x^2+3x+2\right)^2}\right) \\=\ln(x+2)+\ln\left(\frac{\sqrt{x}}{x^2+3 x+2}\right) \\=\ln \frac{(x+2) \sqrt{x}}{(x+1)(x+2)}=\ln \frac{\sqrt{x}}{x+1} . $$
How the TA made me solve it:
$$\frac{1}{3} \ln (x+2)^3+\frac{1}{2}\left[\ln x-\ln \left(x^2+3 x+2\right)^2\right]$$ $$ \begin{aligned} & \ln (x+2)^3+\frac{1}{2}\ln \left(\frac{x}{\left(x^2+3 x+2\right)^2}\right) \\ & \ln (x+2)^3+\ln \left(\sqrt{\frac{x}{\left(x^2+3 x+2\right)^2}}\right) \end{aligned} $$ $$\ln \left[(x+2)^3 \cdot \frac{\sqrt{x}}{\left|x^2+3 x+2\right|}\right] \\\Rightarrow \ln \left(\frac{\sqrt{x}(x+2)^3}{|(x+2)(x+1)|}\right)$$
Your solution seems correct to me .
The second solution contains some mistake and missing points .
The remarkable point here is that, the original domains may change when simplifying logarithms . Therefore, when applying logarithm rules, we generally have to be careful with domains .
We observe that, the original expression includes $\ln x$ . This forces that, we have $x>0$ . This implies that, $x+2>0$ and $x^2+3x+2>0$ or equivalently, $x+1>0$ or $x+2>0$ .
Therefore, we have :
$$ \begin{align}\frac 13\ln (x+2)^3&=3\cdot \frac 13 \ln (x+2)\\&=\ln (x+2)\end{align} $$
and $\require{cancel}$
$$ \begin{align}\ln \frac{\cancel{(x+2)}\sqrt{x}}{(x+1)\cancel{(x+2)}}=\ln \frac{\sqrt{x}}{x+1}\end{align} $$
and
$$ \begin{align}\sqrt \frac{x}{(x^2+3x+2)^2}=\frac {\sqrt x}{(x+1)(x+2)}\end{align} $$
Thus, we do not need to take absolute value here .
As a final note, we see that both original and simplified logarithms have exactly the same domains .