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That is to ask, is $e^{\ln(q_0)}$ = $\ln(e^{q_0})$?

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This condition does not imply that the two maps are inverse, but yes, this is true for sufficiently small q_0. In fact the inverse relationship between the logarithm and the exponential is formal in the sense that it is an equality of power series so it holds whenever the series converge. –  Qiaochu Yuan Feb 24 '11 at 18:52
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2 Answers 2

up vote 2 down vote accepted

$e^{\ln(q_0)} = q_0$ always holds for all branches of the logarithm and even for square matrices (because this is essentially the definition of the log).

$\ln(e^{q_0}) = q_0$ is already for complex numbers not always true (because the logarithm is multivalued)

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I wrote software to play with quaternions numerically on the command line (http://sourceforge.net/projects/quaternions/).

q_ln 1 2 3 4 | q_exp 1.0000000000000011 2.0000000000000000 3.0000000000000004 4.0000000000000000
This confirms $e^{ln(q_0)}=q_0$. Flip the order: q_exp 1 2 3 4 | q_ln 1.0000000000000000 -0.3335164408915938 -0.5002746613373907 -0.6670328817831875 So $\ln(e^{q_0}) \ne q_0$.

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