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:\( \log_{1,b}(-1)=6*I \)
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

