# I Logarithm of negative base to a number resulting in even

1. Jan 7, 2017

### Logical Dog

Negative number multiplied by itself an even number of times gives us a positive number.

Why does log to -10 base of 100 not equal 2?

2. Jan 7, 2017

### Staff: Mentor

You cannot extend that to a function over anything apart from a few selected integers. How would that be useful?

3. Jan 7, 2017

### Staff: Mentor

Logarithmic functions are the inverse of exponentials and are limited to positive reals only for log bases and for log arguments and the base can't be equal to 1. Hence you're not going to see a negative base:

https://en.m.wikipedia.org/wiki/Logarithm

4. Jan 8, 2017

### Svein

Your question only makes sense in the complex domain, so:
$\log_{-10}(100)=\frac{\log(100)}{\log(-10)}=\frac{2\cdot \log(10)}{\log(10)+(2n+1)\pi i}$ where n is an integer (the complex logarithm is not single-valued). Thus, even if you put n=0, the answer is not going to be 2.

5. Jan 8, 2017

### Logical Dog

ok

Last edited: Jan 8, 2017
6. Jan 8, 2017

### Staff: Mentor

Svein's post is a more mathematical version of "you could not extend this to anything useful" - you would run into weird results everywhere.

7. Jan 9, 2017

### Svein

On reflection, my answer is too simple, a more correct answer is $\log_{-10}(100)=\frac{\log(100)}{\log(-10)}=\frac{2\cdot \log(10)+2m\pi i}{\log(10)+(2n+1)\pi i}$ where m and n are integers. The expression can be simplified somewhat, but none of the results are going to be 2.

8. Jan 9, 2017

ok

9. Jan 9, 2017

### Staff: Mentor

Just as a side comment, what you wrote is not clear. It would be clearer as $\log_{-10}(100)$ or in words as "log, base -10, of 100".

10. Jan 11, 2017

### micromass

Staff Emeritus
Take $m=1$ and $n=0$, you do get

$$\frac{2\cdot \log(10)+2\pi i}{\log(10)+\pi i} = 2 \frac{\log(10)+\pi i}{\log(10)+\pi i} = 2$$

So $2$ is a value of $\log_{-10}(100)$, as it should be. But it's not the only value.

11. Jan 11, 2017

### Svein

Talk about missing the obvious. One demerit for me!