Answer
$\log \left(\dfrac{x+2}{x+1} \right)$
Work Step by Step
Recall:
$\log_a M^r = r \log_a M$
Using the rule above gives:
$2 \log (x+1) = \log (x+1)^2$
So the given expression is equivalent to:
$\log (x^2+3x+2) - \log(x+1)^2$
With $\log_a\left(\dfrac{M}{N}\right) = \log_a M-\log_a N$, then the expression above is equivalent to:
$\log \left(\dfrac{x^2+3x+2}{(x+1)^2} \right)$
Note that $ x^2+3x+2=(x+2)(x+1)$.
Hence, the expression above simplifies to:
$\require{cancel}
\log \left(\dfrac{(x+2)(x+1)}{(x+1)^2} \right) \\\\
\log \left(\dfrac{(x+2)\cancel{(x+1)}}{(x+1)^\cancel{2}} \right) \\\\
= \log \left(\dfrac{x+2}{x+1} \right)$
Therefore,
$\log (x^2+3x+2) - 2 \log (x+1) = \boxed{\log \left(\dfrac{x+2}{x+1} \right)}$
