How do you write as a single log for #1/3log_3x + 2/3log_3x #?
1 Answer
Oct 29, 2015
Explanation:
#1/3 log_3(x) + 2/3 log_3(x) = (1/3 + 2/3) log_3(x) = log_3(x)#
#1/3 log_3(x) + 2/3 log_3(x) = (1/3 + 2/3) log_3(x) = log_3(x)#