Question

Still need help please post in this one that way i don't have to search through the old stuff

h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ? i d = 1 2 7 2 3 2 6 8 8 0

Answer 1

Kate, here is a hint how to find your older posts.
Just enter Kate in the search window at the top-right of this page, and it will give you the links to all your previous posts

let me try a different approach to your problem

given log(40√3)/log(4n) = log 45/log (3n)
cross-multiply ...
(log 40√3)(log 3n) = (log 45)(log 4n)
(log 40√3)(log 3 + logn) = (log 45)(log 4 + logn)
(log 40√3)(log3) + (logn)(log40√3) = (log45)(log4) + (logn)(log45)
logn (log40√3 - log45) = (log45)(log4) - (log40√3)(log3)
logn = [(log45)(log4) - (log40√3)(log3)]/(log40√3 - log45)

but log n^3
= 3logn
= 3 [(log45)(log4) - (log40√3)(log3)]/(log40√3 - log45)

(I got appr. -13.532)