SOLUTION: I REALLY REALLY NEED HELP A.S.A.P. What is the value of N in the equation: log{{{8}}}(n-3) + log{{{8}}}(n+4) = 1. I REALLY REALLY appreciate the help because I really need it. Sorr

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Question 20724: I REALLY REALLY NEED HELP A.S.A.P. What is the value of N in the equation: log(n-3) + log(n+4) = 1. I REALLY REALLY appreciate the help because I really need it. Sorry for the short notice.
Found 2 solutions by venugopalramana, kapilsinghi:
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
i think you mean logs to base 8 ...assuming that and using the formula
log x to base y = log x /log y...where all logs are taken to any common base.
so we have
log(n-3)/log 8 + log(n+4)/log 8 =1....multipling through out with log 8,we get
log (n-3)+log (n+4)=log 8
log(n-3)(n+4)=log 8
hence (n-3)(n+4)=8
n^2-3n+4n-12-8=0
n^2+n-20=0
n^2+5n-4n-20=0
n(n+5)-4(n+5)=0
(n+5)(n-4)=0
hence n=-5 or n=4

Answer by kapilsinghi(68) (Show Source):
You can put this solution on YOUR website!
log(n-3) + log(n+4) = 1
log(n-3) + log(n+4) = log(8)
log(n-3)(n+4) = log(8)
taking antilog on both sides
n^2 + n - 12 = 8
n^2 + n - 20 = 0
(n +5)(n-4)=0
n = -5
or n = 4
by substitution n=4 seems correct
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