Imaginary-base logarithm

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An imaginary-base logarithm occurs when one has a logarithm with base i. Though calculating these logarithms may seem to be a daunting task, it can be done easily with the formula

\log_i(x) = {{-2 \ln(x) i} \over \pi} .\,

[edit] Proof of the formula

From the work of Euler, we know that

i^i = e^{{-\pi} / 2}.\,

Using a common identity with e and the natural logarithm, we get

e^{i \ln(i)} = e^{{-\pi} / 2} .\,

Taking the natural logarithm of each side yields

i \ln(i) = {{- \pi} \over 2} .\,

Divide by i:

\ln(i) = {{- \pi} \over {2i}} .\,

By taking the inverse of both sides, we have

\log_i(e) = {-2i \over \pi} .\,

We then multiply each side by the natural logarithm of x

\ln(x)\log_i(e) = {-2\ln(x)i \over \pi} .\,

Using a logarithmic property, we obtain:

\log_i({e^{\ln(x)}}) = {-2\ln(x)i \over \pi} .\,

Once again using the identity with e and the natural logarithm, one obtains the final result of

\log_i(x) = {{-2 \ln(x) i} \over \pi} .\,

Q.E.D.

[edit] Other multiples of i for bases

If the base is a multiple of i, the formula can be generalized as follows:

\log_{ki}(x) = {{-2 \ln(x) i} \over {\pi + 2 \ln(k)}} .\,

This can be proved using the same method above but by adding ln(k) to both sides of the fourth equation shown in the proof.

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