How do you solve #log_3 4x^2-log_3 8=2#?

1 Answer
Cesareo R.
May 24, 2016

#x = 3 sqrt(2)# and #x = -3 sqrt(2)#

Explanation:

#log_3(4x^2)-log_3 8 = log_3(4x^2/8) = log_3 3^2#
so follows
#x^2/2=3^2# having two solutions
#(x/sqrt(2))^2-3^2=(x/sqrt(2)+3)(x/sqrt(2)-3) = 0#
The solutios are
#x = 3 sqrt(2)# and #x = -3 sqrt(2)#

Related questions
See all questions in Logarithmic Models
Impact of this question
1082 views around the world
You can reuse this answer
Creative Commons License