How do you solve log_3 4+log_3 2x=log_3 56?

1 Answer
Feb 24, 2018

#x=7#

Explanation:

Given:

#log_3 4+log_3 2x=log_3 56#

By product rule

#log_b m+log_b n=log_b mn#

Here,
b=3
m=4
n=2x

#log_3 4+log_3 2x=log_3 4xx2x#

Replacing

#log_3 4xx2x=log_3 56#

Bases are same

Equating

#4xx2x=56#

#x=56/(4xx2)#

#56/8=7#

#x=7#