1. ## Simplifying Expressions/Exponents

Can you explain/show the process used to reach this answer?

#1.
(2b^4 c^-2)^5 (3b^-3 c^-4)^-2

= (32 b^20 c^-10) (9 b^6 c^8) = 32 b^26/9c^2 , How did it turn into a
fraction and not, 288 b^26 c^2?

#2 I started these 2 problems and do not knows how to finish them-
Can you show me how to do it?

(3 b^5 c^-2)^3 / 2^-1 b^-3 c = 27 b^15 c^-6 / 2^-1 b^-3 c = ?

(2b^-4 c)-3/ (2b^2 c^-5)^2 = -8b^12c^-3/4b^4c^-10 = ?

2. Originally Posted by Ash
Can you explain/show the process used to reach this answer?

#1.
(2b^4 c^-2)^5 (3b^-3 c^-4)^-2

= (32 b^20 c^-10) (9 b^6 c^8) = 32 b^26/9c^2 , How did it turn into a
fraction and not, 288 b^26 c^2?
the power of c was negative, that's why they put it in a denominator. remember, negative powers mean we take the inverse. so $\displaystyle x^{-a} = \frac 1{x^a}$

$\displaystyle \left(2b^4 c^{-2} \right)^5 \left( 3 b^{-3}c^{-4}\right)^{-2} = \left( 32 b^{20} c^{-10} \right) \left( \frac 19 b^{6}c^8 \right)$

$\displaystyle = \frac {32}9b^{20 + 6}c^{-10 + 8}$

$\displaystyle = \frac {32}9 b^{26}c^{-2}$

$\displaystyle = \frac {32b^{26}}{9c^2}$