How do you simplify #8^2 / 8^4#?

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Answer 1 · 2018-05-02T12:29:16

#1/8^2 or 8^-2#

if you are dividing like terms with powers then you subtract the powers

#x^a/x^b =x^(a-b)#

so 2-4=-2

If you imagine #(8xx8)/(8xx8xx8xx8)# then you can cancel two of

the 8s on the top and bottom, leaving two 8s on the bottom.

Answer 2 · 2018-05-02T12:29:35

#1/64#

#8^2/8^4# is basically #8^2*8^-4#
Using power law.
#8^2*8^-4#=#8^(2-4)#
=#8^-2#
=#1/8^2#
=#1/64#