Can someone please show me (or direct me to) a method for solving these two.
I cannot figure out how you would go about getting the answer.
Thanks for any help.
First try to reduce each term to the same base. $\displaystyle 9= 3^2$ and $\displaystyle 27= 3^3$ so $\displaystyle \frac{(9*3^2)^3}{(3*27)^2}= \frac{(3^2*3^2)^3}{(3*3^3)^2}$$\displaystyle = \frac{(3^{2+2})^3}{(3^{1+3})}= \frac{(3^4)^3}{(3^4)^2}= \frac{3^{12}}{3^8}$.
Can you finish that?
Similarly, $\displaystyle 16= 2^4$, $\displaystyle 4= 2^2$, and $\displaystyle 8= 2^3$ so $\displaystyle \frac{(16*4)^2}{(2*8)^3}= \frac{(2^4*2^2)^2}{(2^1*2^3)^3}$$\displaystyle = \frac{(2^6)^2}{(2^4)^3}= \frac{2^{12}}{2^{12}}$.
Can you finish that?
Here are some hints to start you off:
Get a common base. For example, $\displaystyle 9=3^2$ and $\displaystyle 27=3^3$.
Use the following laws of exponents:
$\displaystyle x^ax^b=x^{a+b}$
$\displaystyle \frac{x^a}{x^b}=x^{a-b}$
$\displaystyle (x^a)^b=x^{ab}$