How do you solve #log_25 n=3/2#?
1 Answer
Nov 21, 2016
We know that if
#n = 25^(3/2)#
#n = (sqrt(25))^3#
#n = (5)^3#
#n = 125#
Hopefully this helps!
We know that if
#n = 25^(3/2)#
#n = (sqrt(25))^3#
#n = (5)^3#
#n = 125#
Hopefully this helps!