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What is the logarithm of log -1?
Assuming you are asking about the natural logarithms (base e):log (-1) = i x pithereforelog (log -1) = log (i x pi) = log i + log pi = (pi/2)i + log pi which is approximately 1.14472989 + 1.57079633 i
Why is log i 0.682i Where i is the imaginary number sqr rt -1?
By Euler's formula, e^ix = cosx + i*sinx Taking natural logarithms, ix = ln(cosx + i*sinx) When x = pi/2, i*pi/2 = ln(i) But ln(i) = log(i)/log(e) where log represents logarithms to base 10. That is, i*pi/2 = log(i)/log(e) And therefore log(i) = i*pi/2*log(e) = i*0.682188 or 0.682*i to three decimal places.
What is the ratio of 100 pi?
It is 100*pi : 1 or approx 314.159:1
What is the common logarithm of -2.4969?
The first of an infinite series of solutions is: log10(-2.4969)=ln(-2.4969)/ln(10)=ln(2.4969)/ln(10) +PI*i/ln(10) = .397 + 1.364*i There are an infinite number of additional solutions of the form: .397 + 1.364*i +2*PI*k/ln(10) where k is any integer greater than 0. I got this number by using the change of base identity and a common, complex log identity, neither of which I'm deriving. If you haven't been taught it yet, i = sqrt(-1).
What is the circumference of a 100 foot radius in terms of pi?
circumference = 2*pi*100
