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 #