# What is the natural log of #-0.9#?

##### 1 Answer

#### Answer:

#### Explanation:

Note that the natural logarithm is supposed to be the inverse of the exponential function

#e^x = -0.9#

Note however that

So there is no real value of

The exponential function

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

"Principal value" ?

Any number of the form

By convention, the principal value of

To find the value of

#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#