hej BO,

How did You find this first value, exactly- where does b come in? Is it used as base of logarithm instead of e?

so log 1,b (z) (-1) is log with base b from -1?

Do I understand correctly?

So the problem is now turned to the properties of continuous application of logarithms with base e^pi/2 on -1?

Which in turn is calculated in each step via normal logarithms by formula:

log b (z) = (ln (z) + 2pik)/ ln (b)?

Regards,

Ivars

bo198214 Wrote:

How did You find this first value, exactly- where does b come in? Is it used as base of logarithm instead of e?

so log 1,b (z) (-1) is log with base b from -1?

Do I understand correctly?

So the problem is now turned to the properties of continuous application of logarithms with base e^pi/2 on -1?

Which in turn is calculated in each step via normal logarithms by formula:

log b (z) = (ln (z) + 2pik)/ ln (b)?

Regards,

Ivars