how do you prove log(b^n) of X = 1/nlog(b) of x??

Answer 1

Do you mean

log (base n) (x) = 1/(log (base x) (n))

Well let log (base n) (x) = a
Therefore x = n^a

Take logs (base x)
So 1 = log (base x) (n^a)
= a * log (base x) (n)
But a = log (base n) (x)
Therefore log (base n) (x) * log (base x) (n) = 1
ie log (base n) (x) = 1/(log (base x) (n)) ............. QED

Answer 2

log x to the baseb^n=1/log b^n to the base x (bsase change rule)
=1/nlogb to the base x

Answer 3

log(b^n) = nlog(b)
that's all i got

Answer 4

I am glad you used the little g in your name.

Maybe I am misreading your question, but it does not make sense to me.

Could you double check the phrasing of it please?

Answer 5

log[b^n](x)=y (eq1)
means (b^n)^y=x=b^(n*y) (eq2)
log[b](x)=z (eq3)
means b^z=x (eq4)
from (eq2) and (eq4) we get z=n*y (eq5)
from eg5 we get y=z/n (eq6)
from eq1,eq3,eq6 we get what you wanted:

log[b^n](x)= y = z/n =(log[b](x))/n