Question

How do you evaluate `log_0.8 (12)`?

Answer 1

`log_0.8 12 approx -11.1359106`

Explanation:

Here we will use the logarithm change of base relationship:

`log_a b = ln b/lna`

Hence, `log_0.8 12 = ln 12/ln 0.8`

`approx 2.48490665/-0.22314355`

`approx -11.1359106`

As a check: `0.8^-11.1359106 approx 11.9999999 approx 12`