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#1 2007-11-05 00:20:21
- EPhillips1989
- Member
- Registered: 2007-11-03
- Posts: 29
logs
hey i was just wondering what happens when you take logs of both sides of equations like the following
e^(3ix)=8e^(-PI/2i)
what happens to the 8??? please help asap xxxxxxx
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#2 2007-11-05 01:01:58
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: logs
Because the 8 is multiplied by the exponent, you can split it into a separate log, using the fact that log(ab) = loga + logb.
So, log[8e^(-π/2i)] = log8 + log(e^-π/2i) = log8 - π/2i.
Why did the vector cross the road?
It wanted to be normal.
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#3 2007-11-05 01:06:20
- EPhillips1989
- Member
- Registered: 2007-11-03
- Posts: 29
Re: logs
i got that far do you know how to find x when
3ix=log8-Pi/2i??
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#4 2007-11-05 01:13:35
- EPhillips1989
- Member
- Registered: 2007-11-03
- Posts: 29
Re: logs
iv got to
i(3x+Pi/2)-log8=0
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