Hello, Jordannn1994!

$\displaystyle 2\;\text{Simplify:}$

$\displaystyle (a)\;\log(d^4) - 2\log\sqrt{d} + 3\log(d^2)$
$\displaystyle \log(d^4) - \log(\sqrt{d})^2 + \log(d^2)^3 \;=\;\log(d^4) - \log(d) + \log(d^6)$

. . $\displaystyle =\; \log\left(\frac{d^4\cdot d^6}{d}\right) \;=\;\log(d^9) \;=\;9\log(d)$

$\displaystyle (b)\;\frac{2\log(x^5) - 3\log(x^2)}{2\log(x^2)}$
$\displaystyle \frac{\log(x^5)^2 - \log(x^2)^3}{\log(x^2)^2} \;=\; \frac{\log(x^{10}) - \log(x^6)}{\log(x^4)} \;=\; \frac{\log(\frac{x^{10}}{x^6})}{\log(x^4)} \;=\;\frac{\log(x^4)}{\log(x^4)} \;=\;1$

$\displaystyle (c)\;2\log(p^5) - \log\sqrt[3]{p^9}$
$\displaystyle \log(p^5)^2 - \log(p^9)^{\frac{1}{3}} \;=\;\log(p^{10}) - \log(p^3) \;=\;\log(p^7) \;=\;7\log(p)$

$\displaystyle \text{3. Factor:}$

$\displaystyle (a)\;5pq + 35pr$
$\displaystyle 5p(q + 7r)$

$\displaystyle (b)\;12ab - 9bc + 6bd$
$\displaystyle 3b(4a-3c+2d)$

$\displaystyle (c)\;5bc - 2bd + 5fc - 2fd$

$\displaystyle b(5c-2d) + f(5c-2d) \;=\; (5c-2d)(b+f)$

And STILL $their \ne they're.$