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

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Answer 1 · 2016-08-04T22:10:58

#x=-27/4#

#"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#