Question

How do you solve `3^b=17`?

Answer 1

`b=2.5789`

Explanation:

Lets take the logarithm of both sides of the equation:

`b*log3=log17`

then divide both sides by `log3`:

`b=log17/log3`

My pocket calculator (HP 15C) reads

`b=2.5789`

Answer 2

Real solution:

`b = ln 17 / ln 3`

Complex solutions:

`b = (ln 17 + 2kpi i)/ ln 3" "` for any integer `k`

Explanation:

Given:

`3^b = 17`

Note that `e^(2kpi i) = 1` for any integer `k`.

So, if `e^a = b` then `a = ln b + 2kpi i` for any integer `k`.

So while we find the real solution by taking the real valued natural log, we can also add any integer multiple of `2pi i` to find all the complex solutions too...

Take natural log of both sides of the given equation to get:

`b ln 3 = ln 17 color(grey)(+ 2kpi i)`

Divide both sides by `ln 3` to get:

`b = (ln 17 color(grey)(+2kpi i))/ln 3`