Solve:

Question

Solve:

volleyballlover55 · Answer

(3^8 *2^5*9^0)^-2 (2^-2/ 3^3)^4

volleyballlover55 · Answer

@helpkay

volleyballlover55 · Answer

show the steps it took you to solve it :)

volleyballlover55 · Answer

?? @Mertsj

mertsj · Answer

$$(3^8\times2^5\times9^0)^{-2}(\frac{2^{-2}}{3^{3}})^4=$$
$$\frac{1}{(3^8\times2^5)^2}\times \frac{1}{(2^2\times3^3)^4}$$

volleyballlover55 · Answer

well actually the second part is divided by right?

mertsj · Answer

$$\frac{1}{3^{16}\times2^{10}}\times\frac{1}{2^8\times3^{12}}=$$

mertsj · Answer

$$\frac{1}{2^{18}\times3^{28}}$$

volleyballlover55 · Answer

wait .. so whats the first step to solving this problem..? make sure to explain it in words as well please :)

mertsj · Answer

$$\frac{2^{-2}}{3^3}=\frac{1}{2^23^3}$$

volleyballlover55 · Answer

first step please?? and explain it in words

mertsj · Answer

The first step was to use this property:
$$a ^{-n}=\frac{1}{a^n}$$

volleyballlover55 · Answer

ok and how did you use it in the problem?

mertsj · Answer

I changed 2^-2 to 1 over 2^2

volleyballlover55 · Answer

look.. it would be awesome.if you can do it like this... 

1. ( you explain what your doing in the step.. 
and then show it working it out in the actual problem..

2.. etc..

mertsj · Answer

That would be excellent practice for you:  study the steps I wrote and figure out how the laws of exponents apply and then write it down.