exponent division question

**fran1942** · December 8th 2010, 11:23 AM

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.

**HallsofIvy** · December 8th 2010, 11:34 AM

First try to reduce each term to the same base. $9= 3^2$ and $27= 3^3$ so $\frac{(9*3^2)^3}{(3*27)^2}= \frac{(3^2*3^2)^3}{(3*3^3)^2}$ $= \frac{(3^{2+2})^3}{(3^{1+3})}= \frac{(3^4)^3}{(3^4)^2}= \frac{3^{12}}{3^8}$ .

Can you finish that?

Similarly, $16= 2^4$ , $4= 2^2$ , and $8= 2^3$ so $\frac{(16*4)^2}{(2*8)^3}= \frac{(2^4*2^2)^2}{(2^1*2^3)^3}$ $= \frac{(2^6)^2}{(2^4)^3}= \frac{2^{12}}{2^{12}}$ .

Can you finish that?

**DrSteve** · December 8th 2010, 11:35 AM

Here are some hints to start you off:

Get a common base. For example, $9=3^2$ and $27=3^3$ .

Use the following laws of exponents:

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

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

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

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