Question

3^n-1=2^n+1
how can i solve this problems, please help?

Answer 1

nasty and tricky the way you typed it

did you mean:
3^(n-1) = 2^(n+1) ??

I sure hope so, and will assume that.

take log of both sides
log [3^(n-1) ] = log [2^(n+1) ]
(n-1) log3 = (n+1) log2
n log3 - log3 = n log2 + log2
n log3 - n log2 = log2 + log3
n(log3 - log2) = log2 + log3
n = (log2+log3)/(log3-log2)
= appr 4.42

it verifies in the original.

Answer 2

I assume it is really this:
3^(n-1)=2^(n+1)
take the log of each side
(n-1)log3=(n+1)log2
now solve for n.