How do you solve #log_10 4+log_10 25#?

1 Answer
Jan 24, 2016

Answer:

#log_10 4 + log_10 25 = 2#

Explanation:

If #a, b, c > 0# then:

#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#