How do you solve #log_9 4+log_9(-3x)=2#?

1 Answer
Aug 4, 2016

Answer:

#x=-27/4#

Explanation:

#"So " log_a(b*c)=log_a b+log_a c#

#"The equation " log _9 4+log _9(-3x)=2 " can be written as:"#

#log_9(4*(-3x))=2#

#log_9(-12x)=2#

#"So " log _a b=x rarr b=a^x#

#"Then:"#

#log_9(-12x)=2#

#-12x=9^2#

#-12x=81#

#x=-81/12#

#x=-27/4#