# Log(-1)= Real numer

1. Nov 14, 2007

Hello guys.

I have a great issue.

I need sollution of this problem.

Log(-1) = Real number.

Please write everything clear and specify all steps.

2. Nov 14, 2007

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

3. Nov 14, 2007

### Math Jeans

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

4. Nov 14, 2007

### JasonRox

Where's robert?

5. Nov 14, 2007

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.

6. Nov 14, 2007

### JasonRox

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

7. Nov 14, 2007

### Gib Z

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

8. Nov 14, 2007

### Kummer

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

Last edited: Nov 15, 2007
9. Nov 15, 2007

### 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).

10. Nov 15, 2007

### AlphaNumeric2

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.

11. Nov 15, 2007

### Gib Z

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

12. Nov 15, 2007

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?????

13. Nov 15, 2007

### christianjb

log(10^x)=x

Thus endeth this boring discussion.

14. Nov 15, 2007

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

15. Nov 15, 2007

### jostpuur

The calculation mistake is in this step:

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

Use more parentheses to avoid these.

16. Nov 15, 2007

ok but

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

Last edited: Nov 15, 2007
17. Nov 15, 2007

### jostpuur

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

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.

18. Nov 15, 2007

### CompuChip

$i^2 = -1$
$(i^2)^2 = (-1)^2 = 1$
$\sqrt{(i^2)^2} = \sqrt{1}$
$i^2 = 1$
$i = 1 \implies 0 = 1 - i \implies 0 = \frac{0}{1 - i} = \frac{1 - i}{1 - i} = 1$
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.

19. Nov 15, 2007

### jostpuur

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

20. Nov 15, 2007

### 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.

21. Nov 15, 2007

### JasonRox

Why is this relevant?

22. Nov 15, 2007

### jostpuur

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.

23. Nov 15, 2007

### Benzsk8

Okay I am only in 11th grade and taking pre-calc so I am not a expert in this. However I dont 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

24. Nov 15, 2007

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