Question

How do you evaluate `log_32 2`?

Answer 1

`1/5`

Explanation:

By the logarithm definition,

`y = log_32 2 equiv 32^y = 2` but

`32^y = (2^5)^y = 2^{5y} = 2 = 2^1`

concluding `5y=1` then `y = 1/5`