How do you simplify #log_2 (4^2*3^4)#?

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Answer 1 · 2017-01-12T20:25:41

#4+4log_2 3#

You would use some properties of logarithms:

1)#log(a*b)=loga+logb#

2)#loga^b=bloga#

then

#log_2(4^2*3^4)=log_2(4^2)+log_2(3^4)##->#property 1)

#=2log_2 4+4log_2 3# #->#property 2)

Since #log_2 4=2#,

the expression becomes

#2*2+4log_2 3=4+4log_2 3#