Represent $\frac{1}{3} \ln(x+2)^3 + \frac{1}{2}[\ln x - \ln (x^2+3x+2)^2]$ as a single logarithm
I am having some trouble trying to find the single logarithm for the following:
$$\frac{1}{3} \ln(x+2)^3 + \frac{1}{2}[\ln x - \ln (x^2+3x+2)^2]$$
I understand that I have to use the addition and subtraction rules but the coefficients are confusing me. These are some of the steps I took:
$$\begin{align} \frac{1}{3} \ln(x+2)^3 + \frac{1}{2} [\ln \frac{x}{(x^2+3x+2)^2}] \end{align}$$
Now from here, if I used the addition rule then what would happen to the coefficients? and also, is there a way for me to simplify the powers $2$ and $3$?
Hints would be appreciated!
Thanks!
$\endgroup$3 Answers
$\begingroup$$$\frac{1}{3} \ln(x+2)^3 + \frac{1}{2}[\ln x - \ln (x^2+3x+2)^2]$$
$$=\ln(x+2)+\frac{1}{2} \ln(x)-\ln(x^2+3x+2)$$
$$=\ln\frac{(x+2)(\sqrt x)}{x^2+3x+2}$$
$$=\ln \frac{\sqrt{x}}{x+1}$$
Properties of $\ln(x)$ that have been used :
$$a \ln(b)= \ln (b^a), ln(a)+ln(b)=\ln(ab)$$
$\endgroup$ 9 $\begingroup$Hints: $a\ln b=\ln b^a$, $\ln a + \ln b = \ln ab$, $x^2+3x+2=(x+1)(x+2)$.
$\endgroup$ $\begingroup$Hint $$ b \cdot \log a = \log a^b$$
$\endgroup$
