How do you solve #log_2x + log_2(x-2) = 3#?

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Answer 1 · 2016-04-21T18:35:39

#x=4#

Use Property: #log_b(xy)=log_bx+log_by and log_by=x iff x^b=y#

#log_2(x(x-2))=3#

#2^3=x^2-2x#

#8=x^2-2x#

#0=x^2-2x-8#

#0=(x-4)(x+2)#

#x-4=0 or x+2=0#

#x=4 or x=-2#

#x=4#