Solve Log(-1) = Real Number Problem

[nejnadusho]

Hello guys.

I have a great issue.

I need sollution of this problem.

Log(-1) = Real number.

Please if you now something about that let me know.
Please write everything clear and specify all steps.
Thank you in advance.

 

[nejnadusho]

$$e^{i\pi}=-1$$

Thank you robert but could you be more specific .
And how can I apply it in this problem?

 

[Math Jeans]

He is basically saying that ln(-1)=i*pi. However, it is not a real number.

 

[JasonRox]

Where's robert?

 

[nejnadusho]

He is basically saying that ln(-1)=i*pi. However, it is not a real number.

Thank you guys.

The point is that I already know all the "wrong" answers and I do not need proof that it is an imaginary number.
This is absolutely clear and defined.
I need somethink inovative and creative.

There is a hint about the DEFINITION for Log(-1)

there should be used some of the logarithmitic properties to be prooved for a real number.

I am stucked now.

If somebody can help me please and thank you.

 

[JasonRox]

Log(-1) = Log (i^2) = Log(i) + Log (i) = 2 Log(i)

 

[Gib Z]

There is nothing innovative or creative about finding a completely rubbish and wrong answer.

 

[Kummer]

Log(-1) = Log (i^2) = Log(i) + Log (i) = 2 Log(i)

That is wrong. $$\log (xy)\not =\log (x) +\log(y)$$ for $$x,y\in \mathbb{C}^{\bold{x}}$$.

 

[CompuChip]

log(-1) is not a real number. It simply isn't, by definition of the function.
But nejnadusho, let us appoint this: if you give me a real solution to $x^2 = -1$ then I'll give you a real expression for log(-1).

 

[AlphaNumeric2]

The point is that I already know all the "wrong" answers and I do not need proof that it is an imaginary number.
This is absolutely clear and defined.
I need somethink inovative and creative.

That's like saying "I need someone to think creatively and help me prove 1+1=3".

If you could prove any statement you liked, maths would be completely meaningless.

 

[Gib Z]

If you could prove any statement you liked, maths would be completely meaningless.

Unless you only liked correct statements, which all mathematicians should =]

 

[nejnadusho]

Ok guys what if I say like this.

Log(-1) = x

=> (Log(-1))/(Log10) = x
=> Log(-1) = x*Log10
=> Log(-1) = Log(10^(x)) -where Log10^(x)= 1 therefore
=> Log(-1) = 1

I know it is crazy but what about that?

 

[christianjb]

log(10^x)=x

Thus endeth this boring discussion.

 

[nejnadusho]

log(10^x)=x

Thus endeth this boring discussion.

Log(10^(x)) = x
x*Log10 = x/x
Log10 = 1
?/

 

[jostpuur]

Ok guys what if I say like this.

Log(-1) = x

=> (Log(-1))/(Log10) = x
=> Log(-1) = x*Log10
=> Log(-1) = Log(10^(x)) -where Log10^(x)= 1 therefore
=> Log(-1) = 1

I know it is crazy but what about that?

The calculation mistake is in this step:

$$\textrm{Log}(10^x) = (\textrm{Log}(10))^x$$

Use more parentheses to avoid these.

 

[nejnadusho]

The calculation mistake is in this step:

$$\textrm{Log}(10^x) = (\textrm{Log}(10))^x$$

Use more parentheses to avoid these.

ok but

Log(10^(x)) = 1
and
(Log(10))^x = 1^x = 1 always
?

 

[jostpuur]

ok but

Log(10^(x)) = 1
and
(Log(10))^x = 1^x = 1 always
?

Your equation

$$\textrm{Log}(10^x)=1$$

is wrong, because

$$\textrm{Log}(10^x) = x$$

and x is not 1 in general.

At this point I would like to know your age.

 

[CompuChip]

nejnadusho, what about this:
$i^2 = -1$
$(i^2)^2 = (-1)^2 = 1$
$\sqrt{(i^2)^2} = \sqrt{1}$
$i^2 = 1$
$i = 1 \implies 0 = 1 - i \implies 0 = \frac{0}{1 - i} = \frac{1 - i}{1 - i} = 1$
so 1 = 0 and from this we can prove that log(-1) = 1 (or any real number you like, or anything else you might want to prove but is not true, for that matter).

And no, there is no error in this calculation, it's just creative thinking.

I doubt nejnadusho reads any of our posts though, unless they actually conclude that log(-1) is real.

 

[jostpuur]

opinion about strategy

CompuChip, your example of mistake is more complicated than the original one given by nejnadusho, and is thus very unlikely going to improve the situation

 

[arildno]

OP is just hung up on the mere term "real".
"Unreal" doesn't exist, or what?

So, clearly, non-real numbers don't exist.

 

[JasonRox]

At this point I would like to know your age.

Why is this relevant?

 

[jostpuur]

Why is this relevant?

The motivation to explain depends on it. Some problems with basic concepts are understandable for a young kid, but I would have less patience when explaining basics to an adult.

 

[Benzsk8]

Okay I am only in 11th grade and taking pre-calc so I am not a expert in this. However I don't see how there is any way to get log(-1) = a real number. Because breaking it down that would be 10^x = -1 , x>= 0/ and all those numbers would be positive. And for nejnadusho second point 1^x always = 1 is true to the work he showed for the second part.
So if I am wrong don't all start yelling at me becaus I am only in 11th but from what I have learned and my friends and I have done there isn't a was to get log(-1)=real number

 

[nejnadusho]

Hello guys.

I just want to thank you.

That is enough for me.

I am sure you are right but I also do not feel guilty because somebody created something that leading to so much missunderstandings.

For me the answer is still not clear.
May be I am an adult, but a kid with an unstopable imagination.

However, At this point it is clear that you are write and Log(-1) = imaginary number.
Enough Logarihms for me.
Thank you for your cooperation.

 

[atqamar]

Glad to see you've accepted the fact that $$log(-1)$$ is an imaginary number. Try finding what that imaginary number is. Use Euler's identity: $$e^{\pi i}=-1$$.