SOLUTION: log8(n-3)+log8(n+4)=1

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Question 559121: log8(n-3)+log8(n+4)=1
Answer by lwsshak3(11628) About Me  (Show Source):
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log8(n-3)+log8(n+4)=1
log8[(n-3)(n+4)]=1
convert to exponential form: base(8) raised to log of number(1)=number(n-3)(n+4)
8^1=n^2+n-12
n^2+n-20=0
(n+5)(n-4)=0
n=-5 (reject, (n+4)>0
or
n=4