Question 37437: Solve for n.
2log n-log (n-1)= log4
Answer by fractalier(6550)   (Show Source):
You can put this solution on YOUR website! From 2log n - log (n-1) = log4, we apply the laws of logarithms to get
log [(n^2) / (n - 1)] = log 4
Thus we can exponentiate both sides and get
(n^2) / (n - 1) = 4
Now cross multiply and solve for n...
n^2 = 4(n - 1)
n^2 = 4n - 4
n^2 - 4n + 4 = 0
(n - 2)^2 = 0
and
n = 2
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