How do you calculate #log10^3 + log10(y^2)#?

1 Answer
Sep 20, 2016

Answer:

#color(green)(4+log_10 y^2)#

Explanation:

Since the default base for #log# is #10#

#log 10^3=log_10 10^3 = 3#

and remembering that #log_b pq =log_b p +log_b q#
#log 10(y^2)=log_10 10 +log_10 (y^2)=1 +log_10 (y^2)#

Therefore
#log 10^3 + log 10(y^2) = 4 + log_10 (y^2)#