How do you approximate #log_3 (18)# given #log_3 2=0.6310# and #log_3 7=1.7712#?
1 Answer
Jan 2, 2017
Explanation:
#log_3(18) = log_3(2*3*3)= log_3(2)+log_3(3)+log_3(3)#
# ~~ 0.6310+1+1 = 2.6310#
