can fraction 1/(2^(i-a)) be expressed in logarithm as -log(i-a)??? if so how?????

Question

owlfred · Answer

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous · Answer

Do you mean i as in the imaginary number?

anonymous · Answer

Oh, so it's like x then.

anonymous · Answer

ya. "i" is like x

anonymous · Answer

is this what the first equation would look like?$$1\over{2^{i-a}}$$

anonymous · Answer

ya

anonymous · Answer

the second one would be$$-\log(i-a)$$ then

anonymous · Answer

actually the first equation is converted as second one .. how did this conversion happen????

anonymous · Answer

That's what I'm figuring out now :) Once I finish I'll help you through it.

anonymous · Answer

k. waiting for ur reply... thanks in advance.......

anonymous · Answer

np(:

anonymous · Answer

is there any relation between fraction and logarithm
???

anonymous · Answer

There can be, I'm trying to remember...

anonymous · Answer

kkk

anonymous · Answer

2a-1

anonymous · Answer

@fltdeckab. acually my question is how $$ 1\div \2( ^\ \i( + \a)) is converted  to -\log_{}(i+a) $$..........