Natural Logarithm of Negative Numbers

[prasannapakkiam]

Well I came across this when someone asked me this question:

(-2)^n = 16

I can clearly see n=4. However, he did this:

ln((-2)^n) = ln(16)
n*ln(-2) = ln(16)
n*ln(2)+n*i*pi = ln(16)

How can I show that n=4 from this?

 

[genneth]

It's because n=4 isn't the only possible solution. Remember that log is a multivalued function (like arcsin) -- so introduce the necessary parametrisation, and see that it can be set to zero.

 

[x-is-y]

Log is a multivalued function since $$e^x$$ is a periodic function. Remember that Euler showed that

$$e^{ix} = \cos{x} + i\sin{x}$$ and hence we have that $$e^x = e^{x + 2\pi i n$$ and more general since $$a^x = e^{\ln{a} x}$$ it's true that $$a^x$$ is a periodic function.

 

[prasannapakkiam]

Can someone please Exemplify? I get what they say; but I am still stuck...

 

[D H]

However, he did this:

ln((-2)^n) = ln(16)
n*ln(-2) = ln(16)
n*ln(2)+n*i*pi = ln(16)

How can I show that n=4 from this?

You can't, because your friend made a mistake. As written, n=4 is not a solution. He should have used

$$\begin{align*}<br /> \ln(-2) &= \text{Ln}(2) + (1 + 2k) \pi i \\<br /> \ln(16) &= \text{Ln}(16) + 2m\pi i<br /> \end{align*}$$

where $\text{Ln}(x)$ is the principal value of the natural logarithm and $k$ and $m$ are abitrary integers.

Applying the above to $n\ln(-2) = \ln(16)$ yields

$$n(\text{Ln}(2) + (1 + 2k) \pi i) = \text{Ln}(16) + 2m\pi i$$

From this you should be able to show that n=4 is but one of infinitely many solutions and also that n=4 is the only pure real solution.

 

[prasannapakkiam]

Thanks for all your input. In the end I see that it is quite a simple problem. However, thanks for putting me on track...