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Exponential formula (complex numbers)?
a^(x+b) + c = 0
Make x the subject.
I got log a (c) + (i*pi/ln(a)) - b = x
Is this correct?
2 Answers
- 1 decade agoFavorite Answer
Not quite. If you plug in your solution, you get a^(c ln(a)) + c instead of 0.
I got the following:
x = ln(-c) / ln(a) + n*i*pi / ln(a) - b,
where n is an integer.
(Assuming of course a, c, and ln(a) are nonzero.)
I don't have time to write up the details. Sorry. :-(
Source(s): Degree in Mathematics - ?Lv 44 years ago
z^4 + z = 0 z * (z^3 + a million) = 0 z * (z + a million) * (z^2 - z + a million) = 0 this is by way of the reality z^3 + a million is the sum of two cubes (a^3 + b^3) and as such it factors into (a + b) * (a^2 - ab + b^2) z = 0 z + a million = 0 z = -a million z^2 - z + a million = 0 z = (a million +/- sqrt(a million - 4)) / 2 z = (a million +/- i * sqrt(3)) / 2 z = 0 , -a million , (a million/2) * (a million + i * sqrt(3)) , (a million/2) * (a million - i * sqrt(3))