How do you solve $Log_3 (x+1) = 2 + Log_3 (x-3)$ ?

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Answer 1 · 2016-05-04T18:34:02

$x=7/2$

$log_3 (x+1)-log_3(x-3)=2$

$log_3 ((x+1)/(x-3))=2$

$3^2 = (x+1)/(x-3)$

$9=(x+1)/(x-3)$

$9(x-3)=x+1$

$9x-27=x+1$

$9x-x=1+27$

$8x=28$

$x=28/8=7/2$

How do you solve Log_3 (x+1) = 2 + Log_3 (x-3)Log_3 (x+1) = 2 + Log_3 (x-3)?