How do you evaluate `log_169 (-13)`?

A log expression in this form is asking the question...

"what power of 169 will give -13?"

OR" What index of 169 will make -13?"

I

For any Real value of `x`, `169^x > 0`, so cannot be equal to `-13`

So to find a value for `log_169(-13)` we need to consider Complex logarithms.

Note that `e^(pi i) = -1`, so `ln(-1) = pi i`

In general, if `x < 0` then `ln(x) = ln abs(x) + pi i`

Use the change of base formula to find:

`log_169 (-13)`

`=ln(-13)/ln(169)`

`=ln(-13)/ln(13^2)`

`=(ln(13)+pi i)/(2 ln(13))`

`=1/2 + pi/(2ln(13))i`

This is the principal value of the Complex logarithm.

Other values that satisfy `169^z = -13` are found by adding multiples of `pi/(ln(13))i`

I think that I could make a compromising answer.

Use that, for `a > 0, a^2=(+-a)^2` and

`log_b a = log a/log b`

Now,

`log_169 (-13).`

`log(-13)/log 169`

`=log(-13)/log((+-13)^2)`

`=(log(-13)/log(+-13))/2`

`=1/2`, for the negative sign, and, for the positive sign,

`=log(13e^(i(2n+1)pi))/(2log 13), n = 0, +-1, +-2, +-3, ..`,

using `-1=`cis ( odd integer multiple of `pi)*`

`=( log 13+ i (2 n + 1) pi) / (2log 13), n = 0, +-1, +-2, +-3, ...`

`=1/2+i(2n+1)pi/(2log 13), n = 0, +-1, +-2, +-3,..`.

If students are not to be burdened, these questions could be

reserved for Extraordinary Talent Examinations, after doing good

home work on the answer....