Your instructor needs to learn to use parentheses.
$-\pi \ln\left(\dfrac{y^2}{x}\right) - 2\ln y + \dfrac{1}{2}\ln \left(x^{20}y^{18}\right) = \ln\left(x^{10+\pi}y^{7-2\pi}\right)$
Here is how I would do it. I would use negative exponents:
$-\pi\ln\left(\dfrac{y^2}{x}\right) = -\pi\ln(x^{-1}y^2) = \ln(x^{-\pi}y^{2\pi})$
This way, everything becomes addition rather than a combination of addition and division.
If you cannot learn to post correctly, then think again.
$ \begin{align*}- \pi \log \left( {\frac{{{y^2}}}{x}} \right) - 2\log (y) + \frac{1}{2}\log \left( {{x^{20}}{y^{18}}} \right)&= - 2\pi \log (y) + \pi \log (x) - 2\log (y) + 10\log (x) + 9\log (y) \\&=(7 - 2\pi )\log (y) + (\pi + 10)\log (x) \end{align*}$