How do you solve #log_2 x=1/3 log_2 27#?
1 Answer
Feb 27, 2016
x = 3
Explanation:
using
#color(blue)" laws of logarithms "#
#• logx^n = nlogx hArr nlogx = logx^n #
#• log_bx = log_by rArr x = y # hence :
#log_2x = log_2 27^(1/3) # now
#27^(1/3) = root(3)27 = 3 #
#rArr log_2x = log_2 3 rArr x = 3 #