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

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Answer 1 · 2016-08-04T22:10:58

$x=-27/4$

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

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

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

$log_9(-12x)=2$

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

$"Then:"$

$log_9(-12x)=2$

$-12x=9^2$

$-12x=81$

$x=-81/12$

$x=-27/4$

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