Simplifying using the rule of logs How would you simplify: $\left[\frac{16}{5}\ln(x + 2) - \frac{1}{5}\ln(x - 3) - \ln x \right]_4^6$ and put it in the form $\ln\frac{m}{n}$. Stating the values of m and n.
Note:
$a\ln b = \ln b^a$
$\ln a - \ln b = \ln \frac{a}{b}$
I just cant get the answer, my answer is always a decimal.

The expression evaluated at $x=6$ is
$$
\frac{16}{5}\ln8-\frac{1}{5}\ln3-\ln6=
\frac{48}{5}\ln2-\frac{1}{5}\ln3-\ln2-\ln3=
\frac{43}{5}\ln2-\frac{6}{5}\ln3
$$
Evaluated at $x=4$ it is
$$
\frac{16}{5}\ln6-\frac{1}{5}\ln1-\ln4=
\frac{16}{5}\ln2+\frac{16}{5}\ln3-2\ln2=
\frac{6}{5}\ln2+\frac{16}{5}\ln3
$$
Thus the difference is
$$
\frac{43}{5}\ln2-\frac{6}{5}\ln3-\frac{6}{5}\ln2-\frac{16}{5}\ln3=
\frac{37}{5}\ln2-\frac{22}{5}\ln3=
\frac{1}{5}\ln\frac{2^{37}}{3^{22}}
$$
There's no way to represent it as $\ln(m/n)$, because this would mean
$$
\left(\frac{m}{n}\right)^{\!5}=\frac{2^{37}}{3^{22}}
$$
and this is impossible with integer $m$ and $n$.
