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.
 A: 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 $n$.
A: Notice, $$\left[\frac{16}{5}\ln(x+2)-\frac{1}{5}\ln(x-3)-\ln (x)\right]_{4}^{6}$$
$$=\frac{1}{5}\left[\ln(x+2)^{16}-\ln(x-3)-\ln (x^5)\right]_{4}^{6}$$
$$=\frac{1}{5}\left[\ln\left(\frac{(x+2)^{16}}{x^5(x-3)}\right)\right]_{4}^{6}$$
$$=\frac{1}{5}\left[\ln\left(\frac{(6+2)^{16}}{6^5(6-3)}\right)-\ln\left(\frac{(4+2)^{16}}{4^5(4-3)}\right)\right]$$
$$=\frac{1}{5}\left[\ln\left(\frac{8^{16}}{3\cdot 6^5}\right)-\ln\left(\frac{6^{16}}{4^5}\right)\right]$$
$$=\frac{1}{5}\ln\left(\frac{8^{16}\cdot 4^{5}}{3\cdot 6^{5}\cdot 6^{16}}\right)$$
$$=\ln\left(\frac{2^{48}\cdot 2^{10}}{3\cdot 3^{21}\cdot 2^{21}}\right)^{1/5}$$
$$=\ln\left(\frac{2^{37}}{3^{22}}\right)^{1/5}$$
$$=\ln\left(\frac{2^{37/5}}{3^{22/5}}\right)$$
now, comparing with $\ln\left(\frac{m}{n}\right)$, one should get 
$$m=2^{37/5}, \ \  \ \ n=2^{22/5} $$
