Logarithm of negative base to a number resulting in even

In summary, 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. Thus, there cannot be a negative base for logarithmic functions and the answer to log to -10 base of 100 cannot equal 2. Additionally, in the complex domain, the expression for log to -10 base of 100 involves integers and does not result in a value of 2. Therefore, log to -10 base of 100 will not equal 2.
  • #1
Logical Dog
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97
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.
 
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  • #2
You cannot extend that to a function over anything apart from a few selected integers. How would that be useful?
 
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  • #3
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
 
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  • #4
Bipolar Demon said:
Why does log to -10 base of 100 not equal 2?
Your question only makes sense in the complex domain, so:
[itex]\log_{-10}(100)=\frac{\log(100)}{\log(-10)}=\frac{2\cdot \log(10)}{\log(10)+(2n+1)\pi i} [/itex] 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.
 
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  • #5
mfb said:
You cannot extend that to a function over anything apart from a few selected integers. How would that be useful?
ok
Svein said:
Your question only makes sense in the complex domain, so:
[itex]\log_{-10}(100)=\frac{\log(100)}{\log(-10)}=\frac{2\cdot \log(10)}{\log(10)+(2n+1)\pi i} [/itex] 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.

:oldconfused::redface:
 
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  • #6
Svein's post is a more mathematical version of "you could not extend this to anything useful" - you would run into weird results everywhere.
 
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  • #7
On reflection, my answer is too simple, a more correct answer is [itex] \log_{-10}(100)=\frac{\log(100)}{\log(-10)}=\frac{2\cdot \log(10)+2m\pi i}{\log(10)+(2n+1)\pi i}[/itex] where m and n are integers. The expression can be simplified somewhat, but none of the results are going to be 2.
 
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  • #8
ok:biggrin::check:
 
  • #9
Bipolar Demon said:
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".
 
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  • #10
Svein said:
On reflection, my answer is too simple, a more correct answer is [itex] \log_{-10}(100)=\frac{\log(100)}{\log(-10)}=\frac{2\cdot \log(10)+2m\pi i}{\log(10)+(2n+1)\pi i}[/itex] 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

[tex]\frac{2\cdot \log(10)+2\pi i}{\log(10)+\pi i} = 2 \frac{\log(10)+\pi i}{\log(10)+\pi i} = 2[/tex]

So ##2## is a value of ##\log_{-10}(100)##, as it should be. But it's not the only value.
 
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  • #11
micromass said:
Take ##m=1## and ##n=0##, you do get

[tex]\frac{2\cdot \log(10)+2\pi i}{\log(10)+\pi i} = 2 \frac{\log(10)+\pi i}{\log(10)+\pi i} = 2[/tex]

So ##2## is a value of ##\log_{-10}(100)##, as it should be. But it's not the only value.
:-pTalk about missing the obvious. One demerit for me!
 

Related to Logarithm of negative base to a number resulting in even

1. What is a logarithm of a negative base?

A logarithm of a negative base is the exponent to which the base must be raised to result in a given number. It is denoted as log-b(x) and is read as "log base -b of x."

2. Can a logarithm of a negative base be defined for all numbers?

No, a logarithm of a negative base can only be defined for numbers that are positive. This is because raising a negative number to any power will result in a positive number, which will then have a defined logarithm.

3. What does it mean to have a logarithm of a negative base to a number resulting in even?

If a logarithm of a negative base to a number results in an even number, it means that the base and the number being raised to that base are opposite signs. For example, log-2(8) results in 3, which is even, because -2 and 8 are opposite signs (-2 is negative and 8 is positive).

4. How do you solve for a logarithm of a negative base to a number resulting in even?

To solve for a logarithm of a negative base to a number resulting in even, you can use the property of logarithms that states logb(x) = y if and only if by = x. In other words, you can rewrite the logarithmic equation as an exponential equation and solve for the unknown variable.

5. Can a logarithm of a negative base be a complex number?

Yes, a logarithm of a negative base can result in a complex number. This is because a negative number raised to a fractional exponent can result in a complex number. For example, log-2(0.125) results in 0.333 + 0.577i, which is a complex number with a real part of 0.333 and an imaginary part of 0.577.

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