SOLUTION: log_2(2m+4)+log_2(m-1)=3

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Question 524329: log_2(2m+4)+log_2(m-1)=3
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
log_2(2m+4)+log_2(m-1)=3
log_2(2m+4)(m-1)=3
(2m+4)(m-1)=2^3
2m^2-2m+4m-4=8
2m^2+2m-4=8
2m^2+2m-12=0
m^2+m-6=0
(m+3)(m-2) = 0
m = {-2, 2}
we can throw out the -2 solution (extraneous) leaving:
m = 2