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) &amp;= \text{Ln}(2) + (1 + 2k) \pi i \\<br /> \ln(16) &amp;= \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...