Question

How do you find the exact value of `log_3 (-9)`?

Answer 1

`=2+ik/ln3pi, k = 1, +-3, +-5... ,+-(2n+1)+....`

Explanation:

`log_b x` is complex of unlimited values, if x < 0.#

Here,

`log_3(-9)`

`=log_3(((+-i)^2)(3^2))`, where `i=sqrt(-1)`#

`=2(log_3(+- )i+log_3 3)`

`=2(ln(+-i)/ln 3+1)`, using `log_b a = log _c a/log_cb`

`=2(ln(e^(ik/2pi)/ln3 +1), k = +-1, +--3, +- 5+ ...+- (2n+1),+....`

`=2+ik/ln3pi, k = 1, +-3, +- ... ++-(2n+1)+....`,

using `ln e^z =z`

Answer 2

`log_3 (-9)` is not defined within Real numbers.
As a Complex value I think the answer is
`color(white)("XXX")log_3(-9) =2+pi/ln(3) i`

Explanation:

For Real numbers the `log` function is only defined when its argument is non-negative.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For Complex Numbers:

It has been a long time since I've done this kind of thing so could someone check my results.

One of the common results of Euler's Formula: `e^(ix)=cos(x)+i * sin(x)`
is that
`color(white)("XXX")e^(ipi)=-1`
from which it follows that `log_e(-1)=ipi`

Also we will need to remember the "change of base" formula for logarithms:
`color(white)("XXX")log_a(p) = (log_b(p))/(log_b(a))`

....and here we go!

`log_3(-9) = log_3(9 xx (-1))`
`color(white)("XXX")=log_3(9) +log_3(-1)`

`color(white)("XXX")=2 +log_3(-1)`

`color(white)("XXX")=2+ (log_e (-1))/(log_e(3)`

`color(white)("XXX")=2+(ipi)/(log_e(3))`

or using the common notation of `ln(x)` for `log_e(x)`
and separating the `i` factor from the second term:
`color(white)("XXX")=2+pi/ln(3) i`