How do you simplify #4^ { 9} \div \frac { 4^ { 5} } { 4^ { - 8} } #?

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Answer 1 · 2017-10-25T20:39:16

Set up the problem as a complex fraction and divide out exponents

As a complex fraction the problem can be rewritten as

#( 4^9/1)/( 4^5/4^-8)#

To simply a complex fraction multiply both the top and the bottom fractions by the inverse io the bottom fraction

# {4^9/1 xx 4^-8/4^5}/{4^5/4^-8 xx 4^-8/4^5}#

The bottom fractions turn into 1 and disappear. leaving

# 4^9/1 xx 4^-8/4^5# Multiplying exponents is the same as adding them so

#( 4^+(9-8))/4^5# This gives

# 4^1/4^5 # = # 1/4^4 # =# 4^-4#