1. ## exponent division question

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.

2. 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?

3. 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:

$\displaystyle x^ax^b=x^{a+b}$

$\displaystyle \frac{x^a}{x^b}=x^{a-b}$

$\displaystyle (x^a)^b=x^{ab}$