How do you solve #Log_3x + log_3(x-8) = log_3(8x)#?

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Answer 1 · 2016-03-22T20:53:19

#x=16#

#log_3 x+log_3 (x-8)-log_3(8x)=0#

#log_3((x(x-8))/(8x))=0#-> use properties#log_b(xy)=log_bx+log_by, and log_b(x/y) = log_b x-log_b y#

#log_3 ((x-8)/8)=0#

#3^0=(x-8)/8#-> use property #y=log_bx iff b^y=x#

#1=(x-8)/8#

#8=x-8#

#16=x#