What is #2401 ^ (-5/4)#?

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Answer 1 · 2016-03-02T12:13:46

#1/16807#

The most important thing to know to begin this problem is that #2401=7^4#.

Hence the expression can be rewritten as

#=(7^4)^(-5/4)#

We can then use the rule

#(a^b)^c=a^(bc)#

by multiplying #4# and #-5/4#:

#=7^(4xx-5/4)#

#=7^-5#

Use the rule regarding negative exponents:

#a^-b=1/a^b#

Move #7^5# to the denominator:

#=1/7^5#

#=1/16807#