Question

How do you evaluate `log_27 (1/3)`?

Answer 1

`log_27 (1/3) = -1/3`

Explanation:

By definition `log` is the inverse of exponentiation and vice versa for any base `b`. That is:

`log_b (b^x) = x` for all Real values of `x`

`b^(log_b x) = x` for all `x > 0`

The change of base formula tells us that for any `a, b, c > 0` with `a != 1` and `c != 1`:

`log_a b = (log_c b) / (log_c a)`

So we find:

`log_27 (1/3) = (log_3 (1/3)) / (log_3 27) = (log_3 3^(-1)) / (log_3 3^3) = -1/3`