SOLUTION: Solve for n.
2log n-log (n-1)= log4
Algebra.Com
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
RELATED QUESTIONS
log4^n=1/2 (answered by ankor@dixie-net.com)
solve for n;... (answered by swincher4391,stanbon)
solve for n... (answered by Fombitz)
Solve for N
Log base 4 (Log base 3 (log base 2... (answered by solver91311)
solve for n: n=... (answered by jim_thompson5910)
solve for n: n=... (answered by jim_thompson5910,josmiceli)
n= log(13/3)/log(1+0.08/12)
solve for... (answered by Alan3354)
In log N = log x + 2log... (answered by stanbon)
log... (answered by mukhopadhyay)