How do you solve #log₃x² - log₃(2x) = 2#?
1 Answer
Jan 15, 2016
x = 18
Explanation:
using the law of logs that : logx - log y = log
#(x/y) #
#rArr log_3 x^2 - log_3 2x = log_3 (x^2/(2x)) = log_3(x/2) # using the relationship :
#log_b a = n rArr a = b^n # then
#log_3 (x/2) = 2 rArr x/2 = 3^2 =9 rArr x = 9 xx 2 = 18 #