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

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Answer 1 · 2016-05-10T11:18:47

#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#