If we know that the power a of 6 is 8, then log6 is the base and the logarithm of 4 =?

1-a/4

If the first n terms and Sn of sequence {an} satisfy log3 (Sn + 2) = n-1, then an =?

If A1 = S1, log3 (S1 + 2) = 1-1 = 0, then S1 + 2 = 1, A1 = S1 = - 1
For n > 1, an = SN-S (n-1), Sn = 3 ^ (n-1) - 2
So an = SN-S (n-1) = (3 ^ (n-1) - 2) - (3 ^ (n-2) - 2) = 2 * 3 ^ (n-2)
Consider that n = 1, A1 = - 1 does not satisfy 2 * 3 (n-2)
So A1 = - 1
When n > 1, an = 2 * 3 ^ (n-2)
(Note: when the formula an = SN-S (n-1) is used, A1 should be calculated separately, that is, A1 = S1 should be calculated separately, and then whether the general term is also satisfied is investigated.)

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