Why logarithm's base is limited? log_x(f(x)) < log_x(g(x)), x<=0

[shahar]

Given:

. . . . .\(\displaystyle \log_x\big(f(x)\big)\, <\, \log_x\big(g(x)\big)\, \mbox{ for }\, x\, \leq\, 0\)

Why x can't less or equal to zero?

 

[pka]

Why x can't less or equal to zero?

Who says that \(\displaystyle x\not<0~?\) See HERE
But it notice the answer in a complex number.
In the complex field \(\displaystyle \log(z)= \log(|z|)+{\bf{i}}\arg(z)\).
But if \(\displaystyle z\in\mathbb{R}^-\) then \(\displaystyle \arg(z)=\pi\).

 

[shahar]

logarithm

Who says that \(\displaystyle x\not<0~?\) See HERE
But it notice the answer in a complex number.
In the complex field \(\displaystyle \log(z)= \log(|z|)+{\bf{i}}\arg(z)\).
But if \(\displaystyle z\in\mathbb{R}^-\) then \(\displaystyle \arg(z)=\pi\).

Why the is no solution in the group of the Real Number?
How I know the solution is Complex?!

 

[pka]

Why the is no solution in the group of the Real Number?
How I know the solution is Complex?!

Please post the original question, giving the entire statement along with any context.

 

[shahar]

Here is the full question with a solution

Please post the original question, giving the entire statement along with any context.

Here are the instructions of the rules of this type of algorithm. (Translated by me from Hebrew).
1.There is no explanation Why the rules are defined in this way.
So I ask of part of staffs to try to understand what isn't mentioned in the instruction.
So if you want I can add the next part is continued by questions and solutions.
2.Why the is no solution in the real number and only in the complex number?

 

[pka]

Here are the instructions of the rules of this type of algorithm. (Translated by me from Hebrew).
1.There is no explanation Why the rules are defined in this way.
So I ask of part of staffs to try to understand what isn't mentioned in the instruction.
So if you want I can add the next part is continued by questions and solutions.
2.Why the is no solution in the real number and only in the complex number?

There is a standard change of base theorem: \(\displaystyle log_b(a)=\dfrac{\log(a)}{\log(b)}\).
You posted this in Pre-Algebra forum, therefore it is reasonable to assume that \(\displaystyle b>0\).
If \(\displaystyle b<0\) then \(\displaystyle \log(b)\) is complex and no longer a real number.
Recall that the complex numbers are not ordered.

 

[Dr.Peterson]

Here are the instructions of the rules of this type of algorithm. (Translated by me from Hebrew).
1.There is no explanation Why the rules are defined in this way.
So I ask of part of staffs to try to understand what isn't mentioned in the instruction.
So if you want I can add the next part is continued by questions and solutions.
2.Why the is no solution in the real number and only in the complex number?

First, in the context of inequalities, we must be dealing with real numbers, as complex numbers are not ordered. Probably in the context of your course, complex logarithms are not even considered (though your course is clearly not pre-algebra).

Now, in that context, logarithms with a negative base are not allowed, because exponentials with a negative base are not well-defined. That is, the log is defined by log_a(x) = y when x = a^y, and if a<0, then a^y behaves erratically -- for example, (-2)² = 4 is positive, (-2)³ = -8 is negative, (-2)^2.5 is complex. Trying to invert such a function would be a nightmare. So we simply don't talk about logs with a negative base.

 

[shahar]

Thanks

Thank now I understand.