What is #625^(1/4)# ?
1 Answer
Jul 12, 2017
Explanation:
First find the prime factorisation of

#625# is not divisible by#2# since it ends with an odd digit. 
#625# is not divisible by#3# since the sum of its digits is not divisible by#3# . That is:#6+2+5 = 13# which is not divisible by#3# . 
#625# is divisible by#5# since it ends with a#5# and we find:
#625 = 5*125 = 5*5*25 = 5*5*5*5 = 5^4#
Hence:
#625^(1/4) = (5^4)^(1/4) = 5^(4*1/4) = 5^1 = 5#