How do you write #16^-9/16^-8# using only positive exponents?

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Answer 1 · 2016-09-05T18:59:15

#(1) / (16)#

We have: #(16^(- 9)) / (16^(- 8))#

Using the laws of exponents:

#= 16^(- 9 - (- 8))#

#= 16^(- 9 + 8)#

#= 16^(- 1)#

#= (1) / (16^(1))#

#= (1) / (16)#

Answer 2 · 2016-09-05T19:54:32

#1/16#

Recall two of the laws of indices: Neither is more important - you can choose which to use.

#x^m/x^n= x^(m-n) if m>n " but " x^m/x^n= 1/x^(n-m) if n>m#

#x^-m= 1/x^m#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)(16^-9)/color(blue)(16^-8) " "larr # get rid of the negative indices

=#color(blue)(16^8)/color(red)(16^9)#

=#1/16#

OR:

#16^-9/16^-8 " "larr # subtract the indices:

=#1/16^(-8-(-9)) = 1/16^(-8+9)#

=#1/16#