Question #0bfd7

1 Answer
Feb 4, 2018

#1/2log(36)+2log(3)+1=log(540)# (assuming #log# means #log_10#)

Explanation:

First, we can use the following identity:
#alog_x(b)=log_x(b^a)#

This gives:
#1/2log(36)+2log(3)+1=log(36^(1/2))+log(3^2)+1=#

#=log(6)+log(9)+1#

Now we can use the multiplication identity:
#log_x(a)+log_x(b)=log_x(a*b)#

#log(6)+log(9)+1=log(6*9)+1=log(54)+1#

I'm unsure if this is what the question is asking for, but we can also bring the #1# into the logaritm. Assuming that #log# means #log_10#, we can rewrite the #1# like so:
#log(54)+1=log(54)+log(10)#

Now we can use the same multiplication identity as before to get:
#=log(54*10)=log(540)#