# 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 #