What is the natural log of #-0.9#?

1 Answer
Mar 17, 2018

Answer:

#ln (-0.9) = ln (0.9) + ipi ~~ -0.10536 + ipi#

Explanation:

Note that the natural logarithm is supposed to be the inverse of the exponential function #e^x#. So the answer to our question is a solution of:

#e^x = -0.9#

Note however that #e^x > 0# for all real values of #x#.

So there is no real value of #x# which is a candidate for the natural logarithm.

The exponential function #e^x# is applicable to complex numbers, so we can look for complex solutions of #e^x = -0.9#.

Note that Euler's identity tells us that:

#e^(ipi) + 1 = 0#

So we find:

#e^(ipi + ln 0.9) = e^(ipi) * e^(ln 0.9) = -1 * 0.9 = -0.9#

In fact the principal value of the complex natural logarithm of #-0.9# is #ln 0.9 + i pi#.

"Principal value" ?

Any number of the form #x = ln 0.9 + (2k+1)pii# (with #k# an integer) will satisfy #e^x = -0.9#.

By convention, the principal value of #ln (r e^(i theta))# is #ln r + i theta# for #theta in (-pi, pi]#.

To find the value of #ln 0.9# we can use:

#ln (1+t) = t-t^2/2+t^3/3-t^4/4+...#

So:

#ln (1-t) = t+t^2/2+t^3/3+t^4/4+...#

and:

#ln (0.9) = ln (1-0.1) = -(0.1+0.01/2+0.001/3+0.0001/4+...)#

#~~ -0.10536#

So:

#ln (-0.9) = ln (-0.9) + ipi ~~ -0.10536+ipi#