Question

Solve:

Answer 1

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

Answer 2

@helpkay

Answer 3

show the steps it took you to solve it :)

Answer 4

?? @Mertsj

Answer 5

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

Answer 6

well actually the second part is divided by right?

Answer 7

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

Answer 8

\[\frac{1}{2^{18}\times3^{28}}\]

Answer 9

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

Answer 10

\[\frac{2^{-2}}{3^3}=\frac{1}{2^23^3}\]

Answer 11

first step please?? and explain it in words

Answer 12

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

Answer 13

ok and how did you use it in the problem?

Answer 14

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

Answer 15

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..

Answer 16

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.