How do you simplify #(7^5 * 7^3)/7^2#?

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Answer 1 · 2016-12-16T14:36:30

The answer is #7^6# or 117,649.

Here are some rules when working with exponents(powers):

#a^m*a^n = a^(m+n)#
#a^m-a^n = a^(m-n)#
#a^0 = 1#

where, #a# is any natural number(1,2,3,4...) and #m and n# are whole numbers(0,1,2,3,4.....) or fractions.

So, in this question,

#(7^5*7^3)/7^2 = 7^(5+3-2) = 7^6 = 117,649#

Answer 2 · 2016-12-16T14:45:55

#7^6 = 117649#

Note that #" "7^5xx7^3" " =" " 7^(5+3)" "=" "7^8#

But #" "7^8# is the same as #7^6xx7^2#

Write #" "(7^5xx7^3)/7^2" as "(7^6xx7^2)/7^2#

#7^6xx7^2/7^2" "=" "7^6xx1" "=" "7^6##