How do you solve ${log}_{6} (x + 2) = {log}_{6} (3 x)$ $log_6 (x + 2) = log_6 (3x)$ ?

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Answer 1 · 2016-05-10T11:18:47

$x = 1$ $x=1$

$color(blue)(log_x(y)=log_x(z)=>y=z, if y>0 and z>0$

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

$x+2=3x and x+2>0 and 3x>0$

$2x=2 and x> -2 and x>0$

$x=1 and x>0$

$x=1$

How do you solve log_6 (x + 2) = log_6 (3x)log6(x+2)=log6(3x)log_6 (x + 2) = log_6 (3x)?