How do you solve #log_4 (x + 2) - log_4 (x - 4) = log_4 3#?

Question

2015 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-12T13:34:24

#log_4(x+2)-log_4(x-4)=log_4(3)#

First calculate the dominion:

#x+2>0 and x-4>0 -> x> -2 and x>4 =>x>4#

Now add log_4(x-4) to both sides of the equality:

#log_4(x+2)=log_4(x-4)+log_4(3)#

We know that log(a)+log(b)=log(ab), so:

#log_4(x+2)=log_4(3(x-4))=log_4(3x-12)#

If a>0, and b>0, log(a)=log(b) => a=b

#x+2=3x-12#

#-2x=-14#

#x=7# (x>4, so the solution is valid)