How do you solve #log_2 (3x)-log_2 7=3#?

1 Answer
Jul 8, 2016

Answer:

Use a property of logs to simplify and solve an algebraic equation to get #x=56/3#.

Explanation:

Begin by simplifying #log_2 3x-log_2 7# using the following property of logs:
#loga-logb=log(a/b)#
Note that this property works with logs of every base, including #2#.

Therefore, #log_2 3x-log_2 7# becomes #log_2 ((3x)/7)#. The problem now reads:
#log_2 ((3x)/7)=3#

We want to get rid of the logarithm, and we do that by raising both sides to the power of #2#:
#log_2 ((3x)/7)=3#
#->2^(log_2 ((3x)/7))=2^3#
#->(3x)/7=8#

Now we just have to solve this equation for #x#:
#(3x)/7=8#
#->3x=56#
#->x=56/3#

Since this fraction cannot be simplified further, it is our final answer.