How do you solve $Log_6 (x+3) = 1- log_6 (x-2)$ ?

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Answer 1 · 2018-06-26T10:03:56

$x=3$

$log_6(x+3)=1+log_6(x-2)$

$log_6(x+3)=log_6 6+log_6(x-2)$

$log_6(x+3)=log_6 6(x-2)$

$x+3=6(x-2)$

$x+3=6x-12$

$5x=15$

$x=3$

Remember:
$log_a b+log_a c=log_a bc$
$log_a a=1$

How do you solve Log_6 (x+3) = 1- log_6 (x-2)Log_6 (x+3) = 1- log_6 (x-2)?