How do you simplify #(b(b^-5)b^-2)/(b(b^8))#?

1 Answer
Nov 9, 2017

Answer:

See a solution process below:

Explanation:

First, cancel common terms in the numerator and denominator:

#(color(red)(cancel(color(black)(b)))(b^-5)b^-2)/(color(red)(cancel(color(black)(b)))(b^8)) => (b^-5b^-2)/b^8#

Next, use this rule for exponents to simplify the numerator:

#x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))#

#(b^color(red)(-5)b^color(blue)(-2))/b^8 => b^(color(red)(-5)+color(blue)(-2))/b^8 => b^-7/b^8#

Now, use this rule for exponents to complete the simplification:

#x^color(red)(a)/x^color(blue)(b) = 1/x^(color(blue)(b)-color(red)(a))#

#b^color(red)(-7)/b^color(blue)(8) => 1/b^(color(blue)(8)-color(red)(-7)) => 1/b^(color(blue)(8)+color(red)(7)) => 1/b^15#