How do you simplify #2log 3+ log4-log6#?

1 Answer
Feb 13, 2016

Answer:

#log(6)#

Explanation:

First, simplify #2log(3)# through the rule:

#b*log(a)=log(a^b)#

Thus, #2log(3)=log(3^2)=log(9)#.

Substitute this back into the original expression:

#=log(9)+log(4)-log(6)#

We can now use the following rule regarding the addition of logarithms (with the same base, which we do have in this scenario):

#log(a)+log(b)=log(ab)#

Thus the first two logarithms can be combined as follows:

#log(9)+log(4)=log(9*4)=log(36)#

Substituting this back into the expression, we obtain:

#=log(36)-log(6)#

Now, to subtract logarithms with the same base, we use the rule:

#log(a)-log(b)=log(a/b)#

Thus, the expression becomes

#=log(36/6)=color(blue)(log(6)#