Question 74025
*[Tex \Large \frac{(x^2)^{-3}(x^{-2})}{(x^2)^{-4}]
When you raise a base with an exponent to an exponent you multiply the exponents (ie *[Tex \Large (x^y)^z=x^{y*z}]). For instance *[Tex \Large (x^2)^{-4}=x^{2*(-4)}=x^{-8}]
*[Tex \Large \frac{(x)^{-3*2}(x^{-2})}{(x)^{-4*2}]
*[Tex \Large \frac{(x)^{-6}(x^{-2})}{(x)^{-8}]
When you multiply like bases with exponents, you add the exponents (ie *[Tex \Large x^2*x^2=x^{2+2}=x^4])
*[Tex \Large \frac{(x)^{-6+(-2)}}{(x)^{-8}]
*[Tex \Large \frac{(x)^{-8}}{(x)^{-8}]Multiply the numerator by adding the exponents
When you divide like bases with exponents, you subtract the exponents (ie *[Tex \Large \frac{x^2}{x^2}=x^{2-2}=x^0=1])
*[Tex \Large (x)^{-8-(-8)}]
*[Tex \Large (x)^{-8+8)}]
*[Tex \Large (x)^{0}=1]
After all of that, we get an answer of 1. This means that for any value of x, I should get 1 everytime
<p>
Check:
Let x equal any number. Lets make x equal to 2 
*[Tex \Large \frac{(x^2)^{-3}(x^{-2})}{(x^2)^{-4}]
*[Tex \Large \frac{(2^2)^{-3}(2^{-2})}{(2^2)^{-4}]Plug in x=2
*[Tex \Large \frac{(4)^{-3}(\frac{1}{4})}{(4)^{-4}]
*[Tex \Large \frac{\frac{1}{4^3}(\frac{1}{4})}{\frac{1}{4^4}]
*[Tex \Large \frac{\frac{1}{64}(\frac{1}{4})}{\frac{1}{256}]
*[Tex \Large \frac{\frac{1}{256}}{\frac{1}{256}]
*[Tex \Large \frac{1}{\not{256}}\frac{\not{256}}{1}=1]These cancel
If you want more proof, let x equal any other number and you will get 1. Hope this makes sense.