Question

How do you solve `log_b(3) = .234`?

Answer 1

`b~~109.39`

Explanation:

If `log_b (3) = .234`
then (by definition of `log`
`color(white)("XXX")b^.234 = 3`

`color(white)("XXX")(b^.234)^(1/.234)=3^(1/.234)`

`color(white)("XXX")b=3^(1/.234)~~3^4.2735~~109.3905`