SOLUTION: log((x+3)(x-8))+ log((x+3)\(x-8)) = 2 I tried but i cannot solve this equation, whose solution is suppose to be x=-13

Question 561221: log((x+3)(x-8))+ log((x+3)\(x-8)) = 2
I tried but i cannot solve this equation, whose solution is suppose to be x=-13
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
log((x+3)(x-8))+ log((x+3)\(x-8)) = 2
log((x+3)+log(x-8))+ log((x+3)-log(x-8)) = 2
log(x-8) cancels out
log(x+3)+log(x+3) = 2
2log(x+3)=2
log[(x+3)^2]=2
convert to exponential form: base(10) raised to log of number(2)=number((x+3)^2)
10^2=(x+3)^2=x^2+6x+9
x^2+6x-91=0
(x+13)(x-7)=0
x+13=0
x=-13
or
x-7=0
x=7 (reject, (x+3)(x-8)>0)
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