PreCalculus
posted by Kate .
Hi i don't know how to do this problem
solve for n^3
given that log(40 SQRT(3))/log(4n) = log(45)/log(3n)
Thanks i'm copletely lost and don't know how to solve

let a=log(40 sqrt(3)) and b=log45
then
a/log(4n)=b/log(3n)
a(log3n)=b*log(4n)
a(log3+logn)=b(log4+logn)
logn(ab)=blog4alog3
logn=( )/(ab)
logn^3=3logn= 3*above
then take the antilog of both sides, it looks like fun. 
hmmm
how did you go from this
a(log3+logn)=b(log4+logn)
to this?
logn(ab)=blog4alog3
logn=( )/(ab) 
You are supposed to fill in the blanks.
alog3+alogn=blon4=blogn
logn(ab)=blog4alog3
logn= (blog4alog3)/(ab) 
well if follow now but don't know how to take the anti log of that mumbo jumbo
n = 10^(3((blog4alog3)/(ab) )
im unsure how that simplifies
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