How do you solve #log_10 4+log_10 25#?
1 Answer
Jan 24, 2016
Explanation:
If
#log_c(c) = 1#
#log_c(a) + log_c(b) = log_c(ab)#
#log_c(a^b) = b log_c(a)#
So we find:
#log_10(4) + log_10(25)#
#=log_10(4 * 25)#
#=log_10(100)#
#=log_10(10^2)#
#=2 log_10(10)#
#=2#
