# 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