How do you solve # log_3 x - log_3 4 = log_3 12#?

2 Answers
Konstantinos Michailidis
Nov 26, 2015

It is

# log_3 x - log_3 4 = log_3 12=>log_3(x/4)=log_3(12)=> x/4=12=>x=48#

Gió
Nov 26, 2015

I found: #x=48#

Explanation:

We can start by using the property of the logs on subtractions to get a ratio of the arguments and write:
#log_3(x/4)=log_3(12)#
raise both side as exponents of #3#:
#3^(log_3(x/4))=3^(log_3(12))# this allows us to cancel out #3# and #log_3# to get:
#cancel(3)^(cancel(log_3)(x/4))=cancel(3)^(cancel(log_3)(12))#
#x/4=12#
#x=48#

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