Logarithm of negative base to a number resulting in even

[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?

thanks in advance.

 

[mfb]

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

 

[jedishrfu]

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

 

[Svein]

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

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.

 

[Logical Dog]

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

ok

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.

 

[mfb]

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

 

[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.

 

[Logical Dog]

ok

 

[Mark44]

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

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".

 

[micromass]

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.

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.

 

[Svein]

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.

Talk about missing the obvious. One demerit for me!